Mixing
Limit without L'Hospital rule
$begingroup$
The value of $$lim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg), xin bigg(0,frac{pi}{2}bigg)$$
Try: Put $displaystyle y=frac{1}{z}$
So $$lim_{zrightarrow 0}lnbigg(frac{sin(x+z)}{sin x}bigg)cdot frac{1}{z}$$
$$ = lim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$$
Using L'Hospital rule
$$lim_{zrightarrow 0}frac{cos(x+z)}{sin (x+z)}=cot x$$
Could some help me? How can I solve without L'Hospital rule? Thanks
limits limits-without-lhopital
edited Jan 5 at 15:10
MJD
47.1k29209392
asked Jan 5 at 9:18
DXTDXT
5,5852630
$endgroup$
add a comment |
$begingroup$
The value of $$lim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg), xin bigg(0,frac{pi}{2}bigg)$$
Try: Put $displaystyle y=frac{1}{z}$
So $$lim_{zrightarrow 0}lnbigg(frac{sin(x+z)}{sin x}bigg)cdot frac{1}{z}$$
$$ = lim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$$
Using L'Hospital rule
$$lim_{zrightarrow 0}frac{cos(x+z)}{sin (x+z)}=cot x$$
Could some help me? How can I solve without L'Hospital rule? Thanks
limits limits-without-lhopital
edited Jan 5 at 15:10
MJD
47.1k29209392
asked Jan 5 at 9:18
DXTDXT
5,5852630
$endgroup$
add a comment |
$begingroup$
The value of $$lim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg), xin bigg(0,frac{pi}{2}bigg)$$
Try: Put $displaystyle y=frac{1}{z}$
So $$lim_{zrightarrow 0}lnbigg(frac{sin(x+z)}{sin x}bigg)cdot frac{1}{z}$$
$$ = lim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$$
Using L'Hospital rule
$$lim_{zrightarrow 0}frac{cos(x+z)}{sin (x+z)}=cot x$$
Could some help me? How can I solve without L'Hospital rule? Thanks
limits limits-without-lhopital
edited Jan 5 at 15:10
MJD
47.1k29209392
asked Jan 5 at 9:18
DXTDXT
5,5852630
$endgroup$
The value of $$lim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg), xin bigg(0,frac{pi}{2}bigg)$$
Try: Put $displaystyle y=frac{1}{z}$
So $$lim_{zrightarrow 0}lnbigg(frac{sin(x+z)}{sin x}bigg)cdot frac{1}{z}$$
$$ = lim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$$
Using L'Hospital rule
$$lim_{zrightarrow 0}frac{cos(x+z)}{sin (x+z)}=cot x$$
Could some help me? How can I solve without L'Hospital rule? Thanks
limits limits-without-lhopital
limits limits-without-lhopital
edited Jan 5 at 15:10
MJD
47.1k29209392
asked Jan 5 at 9:18
DXTDXT
5,5852630
edited Jan 5 at 15:10
MJD
47.1k29209392
asked Jan 5 at 9:18
DXTDXT
5,5852630
edited Jan 5 at 15:10
MJD
47.1k29209392
edited Jan 5 at 15:10
MJD
47.1k29209392
edited Jan 5 at 15:10
MJD
47.1k29209392
47.1k29209392
asked Jan 5 at 9:18
DXTDXT
5,5852630
asked Jan 5 at 9:18
DXTDXT
5,5852630
asked Jan 5 at 9:18
DXTDXT
5,5852630
5,5852630
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
$displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg)=displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x)cos(1/y)+cos(x)sin(1/y)}{sin x}bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(cos(1/y)+cot(x)sin(1/y)bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(1+big[cos(1/y)+cot(x)sin(1/y)-1big]bigg)$
Since $cos(1/y)+cot(x)sin(1/y)-1to0$ as $ytoinfty$, using the standard limit $displaystylelim_{mto0}frac{ln(1+m)}{m}=1$, we get
$=displaystylelim_{yrightarrow infty}ybig(cos(1/y)+cot(x)sin(1/y)-1big)\=displaystylelim_{zto0^+}frac{cos z-1}z+cot(x)cdotlim_{zto0^+}frac{sin z}z$
answered Jan 5 at 9:35
Shubham JohriShubham Johri
4,920717
$endgroup$
$begingroup$
This is how I would have answered but I guess I was late. +1
$endgroup$
– Paramanand Singh
Jan 5 at 19:37
add a comment |
$begingroup$
$displaystylelim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$ is precisely the derivative of $ln(sin x)$ for $xinleft(0,dfrac{pi}{2}right)$using limits.
answered Jan 5 at 9:24
Yadati KiranYadati Kiran
1,749619
$endgroup$
1
$begingroup$
Using the derivative is precisely using L'Hôpital's rule. Which the OP is requesting not to do.
$endgroup$
– mathcounterexamples.net
Jan 5 at 9:26
$begingroup$
@mathcounterexamples.net: You've got to use L'Hopital's rule at some stage.
$endgroup$
– Yadati Kiran
Jan 5 at 9:36
$begingroup$
@mathcounterexamples.net: Not exactly (L'Hospital's Rule involves differentiation followed by a limit operation) but if one forbids L'Hospital's Rule then the implicit intention is to forbid derivatives as well.
$endgroup$
– Paramanand Singh
Jan 5 at 18:58
$begingroup$
You don't necessarily need to use L'Hospital's Rule at some stage. Just use basic limit laws and you get the desired answer.
$endgroup$
– Paramanand Singh
Jan 5 at 19:00
$begingroup$
@ParamanandSingh: Differentiation with limits uses L'Hopital rule implicitly in certain cases. How will one evaluate using just basic limit laws ? Can you elaborate.
$endgroup$
– Yadati Kiran
Jan 5 at 19:32
|
show 2 more comments
$begingroup$
Beside the simple solution given by Yadati Kiran, you could use Taylor series
$$frac{sin(x+z)}{sin (x)}=cot (x) sin (z)+cos (z)$$ Now, using the classical expansion of $sin(z)$ and $cos(z)$ you then have
$$cot (x) sin (z)+cos (z)=1+z cot (x)-frac{z^2}{2}+Oleft(z^3right)$$ Continue with the logarithm to get
$$log (cot (x) sin (z)+cos (z))=z cot (x)-frac{1}{2} z^2 left(cot ^2(x)+1right)+Oleft(z^3right)$$
$$frac 1 z log (cot (x) sin (z)+cos (z))= cot (x)-frac{1}{2} z left(cot ^2(x)+1right)+Oleft(z^2right)$$ which shows the limit and also how it is approached.
answered Jan 5 at 9:37
Claude LeiboviciClaude Leibovici
120k1157132
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add a comment |
$begingroup$
This is not a solution, more of a question.
I get to
1)$log left ( cos (1/y) +frac{sin (1/y)}{1/y}frac{cot x}{y}right )^y.$
$y rightarrow infty : $
Known:
2) $lim_{y rightarrow infty} log (1+(cot x)/y)^y= cot x$.
(Recall $lim_{y rightarrow infty}(1+a/y)^y= e^a$, $a$ real.)
Is there a fairly straightforward way to get from 1)to 2) using limit manipulations ?
Could not come up with a solution.
Thanks
answered Jan 5 at 10:49
Peter SzilasPeter Szilas
11k2821
$endgroup$
$begingroup$
The way is not fairly straightforward but not that difficult either. Use the fact that if $f(y) >0$ and $y(f(y)-1) to 0$ as $ytoinfty $ then ${f(y)} ^yto 1$ as $yto infty$ (prove this!). Use $$f(y) =dfrac{cos(1/y)+sin(1/y)cot x} {1+dfrac{cot x} {y}} $$ and show that $y(f(y) - 1)to 0$ as $ytoinfty $.
$endgroup$
– Paramanand Singh
Jan 5 at 19:23
$begingroup$
Paramand.Thanks a lot. I'll go through your suggestions. Was intrigued by [(cos 1/y)^y](1+[(cos (1/y))^{-1}]sin(1/y)(1/y)^{-1}[(cos x)/y])^y, By multiplication rule can consider lim (cos 1/y)^y×lim(1+A(y)(cos x)/y)^y, where lim A(y)=1, could not handle this second limit .
$endgroup$
– Peter Szilas
Jan 5 at 19:55
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg)=displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x)cos(1/y)+cos(x)sin(1/y)}{sin x}bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(cos(1/y)+cot(x)sin(1/y)bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(1+big[cos(1/y)+cot(x)sin(1/y)-1big]bigg)$
Since $cos(1/y)+cot(x)sin(1/y)-1to0$ as $ytoinfty$, using the standard limit $displaystylelim_{mto0}frac{ln(1+m)}{m}=1$, we get
$=displaystylelim_{yrightarrow infty}ybig(cos(1/y)+cot(x)sin(1/y)-1big)\=displaystylelim_{zto0^+}frac{cos z-1}z+cot(x)cdotlim_{zto0^+}frac{sin z}z$
answered Jan 5 at 9:35
Shubham JohriShubham Johri
4,920717
$endgroup$
$begingroup$
This is how I would have answered but I guess I was late. +1
$endgroup$
– Paramanand Singh
Jan 5 at 19:37
add a comment |
$begingroup$
$displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg)=displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x)cos(1/y)+cos(x)sin(1/y)}{sin x}bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(cos(1/y)+cot(x)sin(1/y)bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(1+big[cos(1/y)+cot(x)sin(1/y)-1big]bigg)$
Since $cos(1/y)+cot(x)sin(1/y)-1to0$ as $ytoinfty$, using the standard limit $displaystylelim_{mto0}frac{ln(1+m)}{m}=1$, we get
$=displaystylelim_{yrightarrow infty}ybig(cos(1/y)+cot(x)sin(1/y)-1big)\=displaystylelim_{zto0^+}frac{cos z-1}z+cot(x)cdotlim_{zto0^+}frac{sin z}z$
answered Jan 5 at 9:35
Shubham JohriShubham Johri
4,920717
$endgroup$
$begingroup$
This is how I would have answered but I guess I was late. +1
$endgroup$
– Paramanand Singh
Jan 5 at 19:37
add a comment |
$begingroup$
$displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg)=displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x)cos(1/y)+cos(x)sin(1/y)}{sin x}bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(cos(1/y)+cot(x)sin(1/y)bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(1+big[cos(1/y)+cot(x)sin(1/y)-1big]bigg)$
Since $cos(1/y)+cot(x)sin(1/y)-1to0$ as $ytoinfty$, using the standard limit $displaystylelim_{mto0}frac{ln(1+m)}{m}=1$, we get
$=displaystylelim_{yrightarrow infty}ybig(cos(1/y)+cot(x)sin(1/y)-1big)\=displaystylelim_{zto0^+}frac{cos z-1}z+cot(x)cdotlim_{zto0^+}frac{sin z}z$
answered Jan 5 at 9:35
Shubham JohriShubham Johri
4,920717
$endgroup$
$displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg)=displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x)cos(1/y)+cos(x)sin(1/y)}{sin x}bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(cos(1/y)+cot(x)sin(1/y)bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(1+big[cos(1/y)+cot(x)sin(1/y)-1big]bigg)$
Since $cos(1/y)+cot(x)sin(1/y)-1to0$ as $ytoinfty$, using the standard limit $displaystylelim_{mto0}frac{ln(1+m)}{m}=1$, we get
$=displaystylelim_{yrightarrow infty}ybig(cos(1/y)+cot(x)sin(1/y)-1big)\=displaystylelim_{zto0^+}frac{cos z-1}z+cot(x)cdotlim_{zto0^+}frac{sin z}z$
answered Jan 5 at 9:35
Shubham JohriShubham Johri
4,920717
edited Jan 5 at 9:40
answered Jan 5 at 9:35
Shubham JohriShubham Johri
4,920717
answered Jan 5 at 9:35
Shubham JohriShubham Johri
4,920717
answered Jan 5 at 9:35
Shubham JohriShubham Johri
4,920717
4,920717
$begingroup$
This is how I would have answered but I guess I was late. +1
$endgroup$
– Paramanand Singh
Jan 5 at 19:37
add a comment |
$begingroup$
This is how I would have answered but I guess I was late. +1
$endgroup$
– Paramanand Singh
Jan 5 at 19:37
$begingroup$
This is how I would have answered but I guess I was late. +1
$endgroup$
– Paramanand Singh
Jan 5 at 19:37
$begingroup$
This is how I would have answered but I guess I was late. +1
$endgroup$
– Paramanand Singh
Jan 5 at 19:37
add a comment |
$begingroup$
$displaystylelim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$ is precisely the derivative of $ln(sin x)$ for $xinleft(0,dfrac{pi}{2}right)$using limits.
answered Jan 5 at 9:24
Yadati KiranYadati Kiran
1,749619
$endgroup$
1
$begingroup$
Using the derivative is precisely using L'Hôpital's rule. Which the OP is requesting not to do.
$endgroup$
– mathcounterexamples.net
Jan 5 at 9:26
$begingroup$
@mathcounterexamples.net: You've got to use L'Hopital's rule at some stage.
$endgroup$
– Yadati Kiran
Jan 5 at 9:36
$begingroup$
@mathcounterexamples.net: Not exactly (L'Hospital's Rule involves differentiation followed by a limit operation) but if one forbids L'Hospital's Rule then the implicit intention is to forbid derivatives as well.
$endgroup$
– Paramanand Singh
Jan 5 at 18:58
$begingroup$
You don't necessarily need to use L'Hospital's Rule at some stage. Just use basic limit laws and you get the desired answer.
$endgroup$
– Paramanand Singh
Jan 5 at 19:00
$begingroup$
@ParamanandSingh: Differentiation with limits uses L'Hopital rule implicitly in certain cases. How will one evaluate using just basic limit laws ? Can you elaborate.
$endgroup$
– Yadati Kiran
Jan 5 at 19:32
|
show 2 more comments
$begingroup$
$displaystylelim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$ is precisely the derivative of $ln(sin x)$ for $xinleft(0,dfrac{pi}{2}right)$using limits.
answered Jan 5 at 9:24
Yadati KiranYadati Kiran
1,749619
$endgroup$
1
$begingroup$
Using the derivative is precisely using L'Hôpital's rule. Which the OP is requesting not to do.
$endgroup$
– mathcounterexamples.net
Jan 5 at 9:26
$begingroup$
@mathcounterexamples.net: You've got to use L'Hopital's rule at some stage.
$endgroup$
– Yadati Kiran
Jan 5 at 9:36
$begingroup$
@mathcounterexamples.net: Not exactly (L'Hospital's Rule involves differentiation followed by a limit operation) but if one forbids L'Hospital's Rule then the implicit intention is to forbid derivatives as well.
$endgroup$
– Paramanand Singh
Jan 5 at 18:58
$begingroup$
You don't necessarily need to use L'Hospital's Rule at some stage. Just use basic limit laws and you get the desired answer.
$endgroup$
– Paramanand Singh
Jan 5 at 19:00
$begingroup$
@ParamanandSingh: Differentiation with limits uses L'Hopital rule implicitly in certain cases. How will one evaluate using just basic limit laws ? Can you elaborate.
$endgroup$
– Yadati Kiran
Jan 5 at 19:32
|
show 2 more comments
$begingroup$
$displaystylelim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$ is precisely the derivative of $ln(sin x)$ for $xinleft(0,dfrac{pi}{2}right)$using limits.
answered Jan 5 at 9:24
Yadati KiranYadati Kiran
1,749619
$endgroup$
$displaystylelim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$ is precisely the derivative of $ln(sin x)$ for $xinleft(0,dfrac{pi}{2}right)$using limits.
answered Jan 5 at 9:24
Yadati KiranYadati Kiran
1,749619
answered Jan 5 at 9:24
Yadati KiranYadati Kiran
1,749619
answered Jan 5 at 9:24
Yadati KiranYadati Kiran
1,749619
answered Jan 5 at 9:24
Yadati KiranYadati Kiran
1,749619
1,749619
1
$begingroup$
Using the derivative is precisely using L'Hôpital's rule. Which the OP is requesting not to do.
$endgroup$
– mathcounterexamples.net
Jan 5 at 9:26
$begingroup$
@mathcounterexamples.net: You've got to use L'Hopital's rule at some stage.
$endgroup$
– Yadati Kiran
Jan 5 at 9:36
$begingroup$
@mathcounterexamples.net: Not exactly (L'Hospital's Rule involves differentiation followed by a limit operation) but if one forbids L'Hospital's Rule then the implicit intention is to forbid derivatives as well.
$endgroup$
– Paramanand Singh
Jan 5 at 18:58
$begingroup$
You don't necessarily need to use L'Hospital's Rule at some stage. Just use basic limit laws and you get the desired answer.
$endgroup$
– Paramanand Singh
Jan 5 at 19:00
$begingroup$
@ParamanandSingh: Differentiation with limits uses L'Hopital rule implicitly in certain cases. How will one evaluate using just basic limit laws ? Can you elaborate.
$endgroup$
– Yadati Kiran
Jan 5 at 19:32
|
show 2 more comments
1
$begingroup$
Using the derivative is precisely using L'Hôpital's rule. Which the OP is requesting not to do.
$endgroup$
– mathcounterexamples.net
Jan 5 at 9:26
$begingroup$
@mathcounterexamples.net: You've got to use L'Hopital's rule at some stage.
$endgroup$
– Yadati Kiran
Jan 5 at 9:36
$begingroup$
@mathcounterexamples.net: Not exactly (L'Hospital's Rule involves differentiation followed by a limit operation) but if one forbids L'Hospital's Rule then the implicit intention is to forbid derivatives as well.
$endgroup$
– Paramanand Singh
Jan 5 at 18:58
$begingroup$
You don't necessarily need to use L'Hospital's Rule at some stage. Just use basic limit laws and you get the desired answer.
$endgroup$
– Paramanand Singh
Jan 5 at 19:00
$begingroup$
@ParamanandSingh: Differentiation with limits uses L'Hopital rule implicitly in certain cases. How will one evaluate using just basic limit laws ? Can you elaborate.
$endgroup$
– Yadati Kiran
Jan 5 at 19:32
1
1
$begingroup$
Using the derivative is precisely using L'Hôpital's rule. Which the OP is requesting not to do.
$endgroup$
– mathcounterexamples.net
Jan 5 at 9:26
$begingroup$
Using the derivative is precisely using L'Hôpital's rule. Which the OP is requesting not to do.
$endgroup$
– mathcounterexamples.net
Jan 5 at 9:26
$begingroup$
@mathcounterexamples.net: You've got to use L'Hopital's rule at some stage.
$endgroup$
– Yadati Kiran
Jan 5 at 9:36
$begingroup$
@mathcounterexamples.net: You've got to use L'Hopital's rule at some stage.
$endgroup$
– Yadati Kiran
Jan 5 at 9:36
$begingroup$
@mathcounterexamples.net: Not exactly (L'Hospital's Rule involves differentiation followed by a limit operation) but if one forbids L'Hospital's Rule then the implicit intention is to forbid derivatives as well.
$endgroup$
– Paramanand Singh
Jan 5 at 18:58
$begingroup$
@mathcounterexamples.net: Not exactly (L'Hospital's Rule involves differentiation followed by a limit operation) but if one forbids L'Hospital's Rule then the implicit intention is to forbid derivatives as well.
$endgroup$
– Paramanand Singh
Jan 5 at 18:58
$begingroup$
You don't necessarily need to use L'Hospital's Rule at some stage. Just use basic limit laws and you get the desired answer.
$endgroup$
– Paramanand Singh
Jan 5 at 19:00
$begingroup$
You don't necessarily need to use L'Hospital's Rule at some stage. Just use basic limit laws and you get the desired answer.
$endgroup$
– Paramanand Singh
Jan 5 at 19:00
$begingroup$
@ParamanandSingh: Differentiation with limits uses L'Hopital rule implicitly in certain cases. How will one evaluate using just basic limit laws ? Can you elaborate.
$endgroup$
– Yadati Kiran
Jan 5 at 19:32
$begingroup$
@ParamanandSingh: Differentiation with limits uses L'Hopital rule implicitly in certain cases. How will one evaluate using just basic limit laws ? Can you elaborate.
$endgroup$
– Yadati Kiran
Jan 5 at 19:32
|
show 2 more comments
$begingroup$
Beside the simple solution given by Yadati Kiran, you could use Taylor series
$$frac{sin(x+z)}{sin (x)}=cot (x) sin (z)+cos (z)$$ Now, using the classical expansion of $sin(z)$ and $cos(z)$ you then have
$$cot (x) sin (z)+cos (z)=1+z cot (x)-frac{z^2}{2}+Oleft(z^3right)$$ Continue with the logarithm to get
$$log (cot (x) sin (z)+cos (z))=z cot (x)-frac{1}{2} z^2 left(cot ^2(x)+1right)+Oleft(z^3right)$$
$$frac 1 z log (cot (x) sin (z)+cos (z))= cot (x)-frac{1}{2} z left(cot ^2(x)+1right)+Oleft(z^2right)$$ which shows the limit and also how it is approached.
answered Jan 5 at 9:37
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
add a comment |
$begingroup$
Beside the simple solution given by Yadati Kiran, you could use Taylor series
$$frac{sin(x+z)}{sin (x)}=cot (x) sin (z)+cos (z)$$ Now, using the classical expansion of $sin(z)$ and $cos(z)$ you then have
$$cot (x) sin (z)+cos (z)=1+z cot (x)-frac{z^2}{2}+Oleft(z^3right)$$ Continue with the logarithm to get
$$log (cot (x) sin (z)+cos (z))=z cot (x)-frac{1}{2} z^2 left(cot ^2(x)+1right)+Oleft(z^3right)$$
$$frac 1 z log (cot (x) sin (z)+cos (z))= cot (x)-frac{1}{2} z left(cot ^2(x)+1right)+Oleft(z^2right)$$ which shows the limit and also how it is approached.
answered Jan 5 at 9:37
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
add a comment |
$begingroup$
Beside the simple solution given by Yadati Kiran, you could use Taylor series
$$frac{sin(x+z)}{sin (x)}=cot (x) sin (z)+cos (z)$$ Now, using the classical expansion of $sin(z)$ and $cos(z)$ you then have
$$cot (x) sin (z)+cos (z)=1+z cot (x)-frac{z^2}{2}+Oleft(z^3right)$$ Continue with the logarithm to get
$$log (cot (x) sin (z)+cos (z))=z cot (x)-frac{1}{2} z^2 left(cot ^2(x)+1right)+Oleft(z^3right)$$
$$frac 1 z log (cot (x) sin (z)+cos (z))= cot (x)-frac{1}{2} z left(cot ^2(x)+1right)+Oleft(z^2right)$$ which shows the limit and also how it is approached.
answered Jan 5 at 9:37
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
Beside the simple solution given by Yadati Kiran, you could use Taylor series
$$frac{sin(x+z)}{sin (x)}=cot (x) sin (z)+cos (z)$$ Now, using the classical expansion of $sin(z)$ and $cos(z)$ you then have
$$cot (x) sin (z)+cos (z)=1+z cot (x)-frac{z^2}{2}+Oleft(z^3right)$$ Continue with the logarithm to get
$$log (cot (x) sin (z)+cos (z))=z cot (x)-frac{1}{2} z^2 left(cot ^2(x)+1right)+Oleft(z^3right)$$
$$frac 1 z log (cot (x) sin (z)+cos (z))= cot (x)-frac{1}{2} z left(cot ^2(x)+1right)+Oleft(z^2right)$$ which shows the limit and also how it is approached.
answered Jan 5 at 9:37
Claude LeiboviciClaude Leibovici
120k1157132
answered Jan 5 at 9:37
Claude LeiboviciClaude Leibovici
120k1157132
answered Jan 5 at 9:37
Claude LeiboviciClaude Leibovici
120k1157132
answered Jan 5 at 9:37
Claude LeiboviciClaude Leibovici
120k1157132
120k1157132
add a comment |
add a comment |
$begingroup$
This is not a solution, more of a question.
I get to
1)$log left ( cos (1/y) +frac{sin (1/y)}{1/y}frac{cot x}{y}right )^y.$
$y rightarrow infty : $
Known:
2) $lim_{y rightarrow infty} log (1+(cot x)/y)^y= cot x$.
(Recall $lim_{y rightarrow infty}(1+a/y)^y= e^a$, $a$ real.)
Is there a fairly straightforward way to get from 1)to 2) using limit manipulations ?
Could not come up with a solution.
Thanks
answered Jan 5 at 10:49
Peter SzilasPeter Szilas
11k2821
$endgroup$
$begingroup$
The way is not fairly straightforward but not that difficult either. Use the fact that if $f(y) >0$ and $y(f(y)-1) to 0$ as $ytoinfty $ then ${f(y)} ^yto 1$ as $yto infty$ (prove this!). Use $$f(y) =dfrac{cos(1/y)+sin(1/y)cot x} {1+dfrac{cot x} {y}} $$ and show that $y(f(y) - 1)to 0$ as $ytoinfty $.
$endgroup$
– Paramanand Singh
Jan 5 at 19:23
$begingroup$
Paramand.Thanks a lot. I'll go through your suggestions. Was intrigued by [(cos 1/y)^y](1+[(cos (1/y))^{-1}]sin(1/y)(1/y)^{-1}[(cos x)/y])^y, By multiplication rule can consider lim (cos 1/y)^y×lim(1+A(y)(cos x)/y)^y, where lim A(y)=1, could not handle this second limit .
$endgroup$
– Peter Szilas
Jan 5 at 19:55
add a comment |
$begingroup$
This is not a solution, more of a question.
I get to
1)$log left ( cos (1/y) +frac{sin (1/y)}{1/y}frac{cot x}{y}right )^y.$
$y rightarrow infty : $
Known:
2) $lim_{y rightarrow infty} log (1+(cot x)/y)^y= cot x$.
(Recall $lim_{y rightarrow infty}(1+a/y)^y= e^a$, $a$ real.)
Is there a fairly straightforward way to get from 1)to 2) using limit manipulations ?
Could not come up with a solution.
Thanks
answered Jan 5 at 10:49
Peter SzilasPeter Szilas
11k2821
$endgroup$
$begingroup$
The way is not fairly straightforward but not that difficult either. Use the fact that if $f(y) >0$ and $y(f(y)-1) to 0$ as $ytoinfty $ then ${f(y)} ^yto 1$ as $yto infty$ (prove this!). Use $$f(y) =dfrac{cos(1/y)+sin(1/y)cot x} {1+dfrac{cot x} {y}} $$ and show that $y(f(y) - 1)to 0$ as $ytoinfty $.
$endgroup$
– Paramanand Singh
Jan 5 at 19:23
$begingroup$
Paramand.Thanks a lot. I'll go through your suggestions. Was intrigued by [(cos 1/y)^y](1+[(cos (1/y))^{-1}]sin(1/y)(1/y)^{-1}[(cos x)/y])^y, By multiplication rule can consider lim (cos 1/y)^y×lim(1+A(y)(cos x)/y)^y, where lim A(y)=1, could not handle this second limit .
$endgroup$
– Peter Szilas
Jan 5 at 19:55
add a comment |
$begingroup$
This is not a solution, more of a question.
I get to
1)$log left ( cos (1/y) +frac{sin (1/y)}{1/y}frac{cot x}{y}right )^y.$
$y rightarrow infty : $
Known:
2) $lim_{y rightarrow infty} log (1+(cot x)/y)^y= cot x$.
(Recall $lim_{y rightarrow infty}(1+a/y)^y= e^a$, $a$ real.)
Is there a fairly straightforward way to get from 1)to 2) using limit manipulations ?
Could not come up with a solution.
Thanks
answered Jan 5 at 10:49
Peter SzilasPeter Szilas
11k2821
$endgroup$
This is not a solution, more of a question.
I get to
1)$log left ( cos (1/y) +frac{sin (1/y)}{1/y}frac{cot x}{y}right )^y.$
$y rightarrow infty : $
Known:
2) $lim_{y rightarrow infty} log (1+(cot x)/y)^y= cot x$.
(Recall $lim_{y rightarrow infty}(1+a/y)^y= e^a$, $a$ real.)
Is there a fairly straightforward way to get from 1)to 2) using limit manipulations ?
Could not come up with a solution.
Thanks
answered Jan 5 at 10:49
Peter SzilasPeter Szilas
11k2821
edited Jan 5 at 12:15
answered Jan 5 at 10:49
Peter SzilasPeter Szilas
11k2821
answered Jan 5 at 10:49
Peter SzilasPeter Szilas
11k2821
answered Jan 5 at 10:49
Peter SzilasPeter Szilas
11k2821
11k2821
$begingroup$
The way is not fairly straightforward but not that difficult either. Use the fact that if $f(y) >0$ and $y(f(y)-1) to 0$ as $ytoinfty $ then ${f(y)} ^yto 1$ as $yto infty$ (prove this!). Use $$f(y) =dfrac{cos(1/y)+sin(1/y)cot x} {1+dfrac{cot x} {y}} $$ and show that $y(f(y) - 1)to 0$ as $ytoinfty $.
$endgroup$
– Paramanand Singh
Jan 5 at 19:23
$begingroup$
Paramand.Thanks a lot. I'll go through your suggestions. Was intrigued by [(cos 1/y)^y](1+[(cos (1/y))^{-1}]sin(1/y)(1/y)^{-1}[(cos x)/y])^y, By multiplication rule can consider lim (cos 1/y)^y×lim(1+A(y)(cos x)/y)^y, where lim A(y)=1, could not handle this second limit .
$endgroup$
– Peter Szilas
Jan 5 at 19:55
add a comment |
$begingroup$
The way is not fairly straightforward but not that difficult either. Use the fact that if $f(y) >0$ and $y(f(y)-1) to 0$ as $ytoinfty $ then ${f(y)} ^yto 1$ as $yto infty$ (prove this!). Use $$f(y) =dfrac{cos(1/y)+sin(1/y)cot x} {1+dfrac{cot x} {y}} $$ and show that $y(f(y) - 1)to 0$ as $ytoinfty $.
$endgroup$
– Paramanand Singh
Jan 5 at 19:23
$begingroup$
Paramand.Thanks a lot. I'll go through your suggestions. Was intrigued by [(cos 1/y)^y](1+[(cos (1/y))^{-1}]sin(1/y)(1/y)^{-1}[(cos x)/y])^y, By multiplication rule can consider lim (cos 1/y)^y×lim(1+A(y)(cos x)/y)^y, where lim A(y)=1, could not handle this second limit .
$endgroup$
– Peter Szilas
Jan 5 at 19:55
$begingroup$
The way is not fairly straightforward but not that difficult either. Use the fact that if $f(y) >0$ and $y(f(y)-1) to 0$ as $ytoinfty $ then ${f(y)} ^yto 1$ as $yto infty$ (prove this!). Use $$f(y) =dfrac{cos(1/y)+sin(1/y)cot x} {1+dfrac{cot x} {y}} $$ and show that $y(f(y) - 1)to 0$ as $ytoinfty $.
$endgroup$
– Paramanand Singh
Jan 5 at 19:23
$begingroup$
The way is not fairly straightforward but not that difficult either. Use the fact that if $f(y) >0$ and $y(f(y)-1) to 0$ as $ytoinfty $ then ${f(y)} ^yto 1$ as $yto infty$ (prove this!). Use $$f(y) =dfrac{cos(1/y)+sin(1/y)cot x} {1+dfrac{cot x} {y}} $$ and show that $y(f(y) - 1)to 0$ as $ytoinfty $.
$endgroup$
– Paramanand Singh
Jan 5 at 19:23
$begingroup$
Paramand.Thanks a lot. I'll go through your suggestions. Was intrigued by [(cos 1/y)^y](1+[(cos (1/y))^{-1}]sin(1/y)(1/y)^{-1}[(cos x)/y])^y, By multiplication rule can consider lim (cos 1/y)^y×lim(1+A(y)(cos x)/y)^y, where lim A(y)=1, could not handle this second limit .
$endgroup$
– Peter Szilas
Jan 5 at 19:55
$begingroup$
Paramand.Thanks a lot. I'll go through your suggestions. Was intrigued by [(cos 1/y)^y](1+[(cos (1/y))^{-1}]sin(1/y)(1/y)^{-1}[(cos x)/y])^y, By multiplication rule can consider lim (cos 1/y)^y×lim(1+A(y)(cos x)/y)^y, where lim A(y)=1, could not handle this second limit .
$endgroup$
– Peter Szilas
Jan 5 at 19:55
add a comment |
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