Mixing
Using Partial Limit
$begingroup$
$$lim_{x to 0} cos(pi/2cos(x))/x^2$$
I tried to evaluate the limit this way,
$$lim_{x to 0} cos(pi/2cdot1)/x^2$$ since $cos0=1$
$$lim_{x to 0} cos(pi/2cdot1)/x^2=lim_{x to 0} 0/x^2$$
Now apply L'Hospital's Rule twice,
$$lim_{x to 0} 0/2(x)=lim_{x to 0} 0/2=0$$
So,this way the answer is zero.
Can you please explain where am I doing wrong?
I will be thankful for help!
calculus limits
edited Jan 20 at 4:29
Larry
2,41331129
asked Jan 20 at 4:27
Navneet KumarNavneet Kumar
4361516
$endgroup$
add a comment |
$begingroup$
$$lim_{x to 0} cos(pi/2cos(x))/x^2$$
I tried to evaluate the limit this way,
$$lim_{x to 0} cos(pi/2cdot1)/x^2$$ since $cos0=1$
$$lim_{x to 0} cos(pi/2cdot1)/x^2=lim_{x to 0} 0/x^2$$
Now apply L'Hospital's Rule twice,
$$lim_{x to 0} 0/2(x)=lim_{x to 0} 0/2=0$$
So,this way the answer is zero.
Can you please explain where am I doing wrong?
I will be thankful for help!
calculus limits
edited Jan 20 at 4:29
Larry
2,41331129
asked Jan 20 at 4:27
Navneet KumarNavneet Kumar
4361516
$endgroup$
3
$begingroup$
You can't simply plug 0 into parts. Take $limlimits_{xto0}frac xx$ for example. Plugging 0 into the numerator implies the limit should be 0, but clearly the limit is 1.
$endgroup$
– Simply Beautiful Art
Jan 20 at 4:49
$begingroup$
@SimplyBeautifulArt But,in some of the questions we do replace 'x' by the constant values such that it simplifies to a non zero number.Why are we allowed to do that?
$endgroup$
– Navneet Kumar
Jan 20 at 4:57
$begingroup$
Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
$endgroup$
– Simply Beautiful Art
Jan 20 at 4:58
$begingroup$
@SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
$endgroup$
– Navneet Kumar
Jan 20 at 5:01
1
$begingroup$
"The only thing that I need to keep in mind is that the expression doesn't become indeterminate." This is vague, and also the reason we have clear limit rules to follow.
$endgroup$
– Simply Beautiful Art
Jan 20 at 5:06
add a comment |
$begingroup$
$$lim_{x to 0} cos(pi/2cos(x))/x^2$$
I tried to evaluate the limit this way,
$$lim_{x to 0} cos(pi/2cdot1)/x^2$$ since $cos0=1$
$$lim_{x to 0} cos(pi/2cdot1)/x^2=lim_{x to 0} 0/x^2$$
Now apply L'Hospital's Rule twice,
$$lim_{x to 0} 0/2(x)=lim_{x to 0} 0/2=0$$
So,this way the answer is zero.
Can you please explain where am I doing wrong?
I will be thankful for help!
calculus limits
edited Jan 20 at 4:29
Larry
2,41331129
asked Jan 20 at 4:27
Navneet KumarNavneet Kumar
4361516
$endgroup$
$$lim_{x to 0} cos(pi/2cos(x))/x^2$$
I tried to evaluate the limit this way,
$$lim_{x to 0} cos(pi/2cdot1)/x^2$$ since $cos0=1$
$$lim_{x to 0} cos(pi/2cdot1)/x^2=lim_{x to 0} 0/x^2$$
Now apply L'Hospital's Rule twice,
$$lim_{x to 0} 0/2(x)=lim_{x to 0} 0/2=0$$
So,this way the answer is zero.
Can you please explain where am I doing wrong?
I will be thankful for help!
calculus limits
calculus limits
edited Jan 20 at 4:29
Larry
2,41331129
asked Jan 20 at 4:27
Navneet KumarNavneet Kumar
4361516
edited Jan 20 at 4:29
Larry
2,41331129
asked Jan 20 at 4:27
Navneet KumarNavneet Kumar
4361516
edited Jan 20 at 4:29
Larry
2,41331129
edited Jan 20 at 4:29
Larry
2,41331129
edited Jan 20 at 4:29
Larry
2,41331129
2,41331129
asked Jan 20 at 4:27
Navneet KumarNavneet Kumar
4361516
asked Jan 20 at 4:27
Navneet KumarNavneet Kumar
4361516
asked Jan 20 at 4:27
Navneet KumarNavneet Kumar
4361516
4361516
3
$begingroup$
You can't simply plug 0 into parts. Take $limlimits_{xto0}frac xx$ for example. Plugging 0 into the numerator implies the limit should be 0, but clearly the limit is 1.
$endgroup$
– Simply Beautiful Art
Jan 20 at 4:49
$begingroup$
@SimplyBeautifulArt But,in some of the questions we do replace 'x' by the constant values such that it simplifies to a non zero number.Why are we allowed to do that?
$endgroup$
– Navneet Kumar
Jan 20 at 4:57
$begingroup$
Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
$endgroup$
– Simply Beautiful Art
Jan 20 at 4:58
$begingroup$
@SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
$endgroup$
– Navneet Kumar
Jan 20 at 5:01
1
$begingroup$
"The only thing that I need to keep in mind is that the expression doesn't become indeterminate." This is vague, and also the reason we have clear limit rules to follow.
$endgroup$
– Simply Beautiful Art
Jan 20 at 5:06
add a comment |
3
$begingroup$
You can't simply plug 0 into parts. Take $limlimits_{xto0}frac xx$ for example. Plugging 0 into the numerator implies the limit should be 0, but clearly the limit is 1.
$endgroup$
– Simply Beautiful Art
Jan 20 at 4:49
$begingroup$
@SimplyBeautifulArt But,in some of the questions we do replace 'x' by the constant values such that it simplifies to a non zero number.Why are we allowed to do that?
$endgroup$
– Navneet Kumar
Jan 20 at 4:57
$begingroup$
Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
$endgroup$
– Simply Beautiful Art
Jan 20 at 4:58
$begingroup$
@SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
$endgroup$
– Navneet Kumar
Jan 20 at 5:01
1
$begingroup$
"The only thing that I need to keep in mind is that the expression doesn't become indeterminate." This is vague, and also the reason we have clear limit rules to follow.
$endgroup$
– Simply Beautiful Art
Jan 20 at 5:06
3
3
$begingroup$
You can't simply plug 0 into parts. Take $limlimits_{xto0}frac xx$ for example. Plugging 0 into the numerator implies the limit should be 0, but clearly the limit is 1.
$endgroup$
– Simply Beautiful Art
Jan 20 at 4:49
$begingroup$
You can't simply plug 0 into parts. Take $limlimits_{xto0}frac xx$ for example. Plugging 0 into the numerator implies the limit should be 0, but clearly the limit is 1.
$endgroup$
– Simply Beautiful Art
Jan 20 at 4:49
$begingroup$
@SimplyBeautifulArt But,in some of the questions we do replace 'x' by the constant values such that it simplifies to a non zero number.Why are we allowed to do that?
$endgroup$
– Navneet Kumar
Jan 20 at 4:57
$begingroup$
@SimplyBeautifulArt But,in some of the questions we do replace 'x' by the constant values such that it simplifies to a non zero number.Why are we allowed to do that?
$endgroup$
– Navneet Kumar
Jan 20 at 4:57
$begingroup$
Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
$endgroup$
– Simply Beautiful Art
Jan 20 at 4:58
$begingroup$
Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
$endgroup$
– Simply Beautiful Art
Jan 20 at 4:58
$begingroup$
@SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
$endgroup$
– Navneet Kumar
Jan 20 at 5:01
$begingroup$
@SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
$endgroup$
– Navneet Kumar
Jan 20 at 5:01
1
1
$begingroup$
"The only thing that I need to keep in mind is that the expression doesn't become indeterminate." This is vague, and also the reason we have clear limit rules to follow.
$endgroup$
– Simply Beautiful Art
Jan 20 at 5:06
$begingroup$
"The only thing that I need to keep in mind is that the expression doesn't become indeterminate." This is vague, and also the reason we have clear limit rules to follow.
$endgroup$
– Simply Beautiful Art
Jan 20 at 5:06
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
When both the numerator and denominator are zero, you have to differentiate both of them to obtain the correct limit. You can't straightly make the numerator $0$. Think about it, if you have
$$lim_{xrightarrow0}frac{2x}{3x}$$
, and you make the numerator zero and differentiate the denominator, you will get zero, which clearly isn't correct.
What I suggest is that you differentiate both the numerator and the denominator to get
$$lim_{xrightarrow0}frac{pisin(x)sin(pi/2cos(x))}{4x}$$
Now differentiate again, and you get
$$lim_{xrightarrow0}frac{pileft[cos(x)sin(pi/2cos(x))-pi/2sin(x)(cos(pi/2cos(x)))right]}{4}=frac{pi}{4}$$
answered Jan 20 at 4:47
LarryLarry
2,41331129
$endgroup$
$begingroup$
Thanks So Much! Please go through this once artofproblemsolving.com/community/q2h1766346p11567701 I just want to know how to decide which 'x' are safe to replace by their constant values?
$endgroup$
– Navneet Kumar
Jan 20 at 5:03
$begingroup$
Nikolas did a pretty good job explaining when it is safe to replace the values. Yes, when the denominator becomes zero, you don't want to replace the value. In this question, the denominator is zero, so you can't just replace the value. You would instead do L'Hospital or some other manipulations as presented in other answers posted here.
$endgroup$
– Larry
Jan 20 at 5:48
add a comment |
$begingroup$
Hint
Compose Taylor series
$$cos(x)=1-frac{x^2}{2}+Oleft(x^4right)$$
$$cos left(frac{pi}{2} cos (x)right)=sin left(frac{pi }{4}x^2+Oleft(x^4right)right)$$ The next step is simple.
answered Jan 20 at 4:43
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
add a comment |
$begingroup$
Using $frac{1-cos(x)}{2}=sin^2(x/2)$ and $lim_{tto 0}frac{sin(t)}{t}=1$, we can write
$$begin{align}
frac{cosleft(fracpi2 cos(x)right)}{x^2}&=frac{sinleft(pi sin^2(x/2)right)}{x^2}\\
&=fracpi4underbrace{left(frac{sin(x/2)}{x/2}right)^2}_{to1,,text{as},,xto0}underbrace{left(frac{sinleft(pi sin^2(x/2)right)}{pisin^2(x/2)}right)}_{to 1,,text{as},,xto0}\\
end{align}$$
Therefore, we find that
$$lim_{xto0}frac{cosleft(fracpi2 cos(x)right)}{x^2}=fracpi4$$
answered Jan 20 at 5:19
Mark ViolaMark Viola
133k1277175
$endgroup$
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
$endgroup$
– Mark Viola
Jan 30 at 5:07
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When both the numerator and denominator are zero, you have to differentiate both of them to obtain the correct limit. You can't straightly make the numerator $0$. Think about it, if you have
$$lim_{xrightarrow0}frac{2x}{3x}$$
, and you make the numerator zero and differentiate the denominator, you will get zero, which clearly isn't correct.
What I suggest is that you differentiate both the numerator and the denominator to get
$$lim_{xrightarrow0}frac{pisin(x)sin(pi/2cos(x))}{4x}$$
Now differentiate again, and you get
$$lim_{xrightarrow0}frac{pileft[cos(x)sin(pi/2cos(x))-pi/2sin(x)(cos(pi/2cos(x)))right]}{4}=frac{pi}{4}$$
answered Jan 20 at 4:47
LarryLarry
2,41331129
$endgroup$
$begingroup$
Thanks So Much! Please go through this once artofproblemsolving.com/community/q2h1766346p11567701 I just want to know how to decide which 'x' are safe to replace by their constant values?
$endgroup$
– Navneet Kumar
Jan 20 at 5:03
$begingroup$
Nikolas did a pretty good job explaining when it is safe to replace the values. Yes, when the denominator becomes zero, you don't want to replace the value. In this question, the denominator is zero, so you can't just replace the value. You would instead do L'Hospital or some other manipulations as presented in other answers posted here.
$endgroup$
– Larry
Jan 20 at 5:48
add a comment |
$begingroup$
When both the numerator and denominator are zero, you have to differentiate both of them to obtain the correct limit. You can't straightly make the numerator $0$. Think about it, if you have
$$lim_{xrightarrow0}frac{2x}{3x}$$
, and you make the numerator zero and differentiate the denominator, you will get zero, which clearly isn't correct.
What I suggest is that you differentiate both the numerator and the denominator to get
$$lim_{xrightarrow0}frac{pisin(x)sin(pi/2cos(x))}{4x}$$
Now differentiate again, and you get
$$lim_{xrightarrow0}frac{pileft[cos(x)sin(pi/2cos(x))-pi/2sin(x)(cos(pi/2cos(x)))right]}{4}=frac{pi}{4}$$
answered Jan 20 at 4:47
LarryLarry
2,41331129
$endgroup$
$begingroup$
Thanks So Much! Please go through this once artofproblemsolving.com/community/q2h1766346p11567701 I just want to know how to decide which 'x' are safe to replace by their constant values?
$endgroup$
– Navneet Kumar
Jan 20 at 5:03
$begingroup$
Nikolas did a pretty good job explaining when it is safe to replace the values. Yes, when the denominator becomes zero, you don't want to replace the value. In this question, the denominator is zero, so you can't just replace the value. You would instead do L'Hospital or some other manipulations as presented in other answers posted here.
$endgroup$
– Larry
Jan 20 at 5:48
add a comment |
$begingroup$
When both the numerator and denominator are zero, you have to differentiate both of them to obtain the correct limit. You can't straightly make the numerator $0$. Think about it, if you have
$$lim_{xrightarrow0}frac{2x}{3x}$$
, and you make the numerator zero and differentiate the denominator, you will get zero, which clearly isn't correct.
What I suggest is that you differentiate both the numerator and the denominator to get
$$lim_{xrightarrow0}frac{pisin(x)sin(pi/2cos(x))}{4x}$$
Now differentiate again, and you get
$$lim_{xrightarrow0}frac{pileft[cos(x)sin(pi/2cos(x))-pi/2sin(x)(cos(pi/2cos(x)))right]}{4}=frac{pi}{4}$$
answered Jan 20 at 4:47
LarryLarry
2,41331129
$endgroup$
When both the numerator and denominator are zero, you have to differentiate both of them to obtain the correct limit. You can't straightly make the numerator $0$. Think about it, if you have
$$lim_{xrightarrow0}frac{2x}{3x}$$
, and you make the numerator zero and differentiate the denominator, you will get zero, which clearly isn't correct.
What I suggest is that you differentiate both the numerator and the denominator to get
$$lim_{xrightarrow0}frac{pisin(x)sin(pi/2cos(x))}{4x}$$
Now differentiate again, and you get
$$lim_{xrightarrow0}frac{pileft[cos(x)sin(pi/2cos(x))-pi/2sin(x)(cos(pi/2cos(x)))right]}{4}=frac{pi}{4}$$
answered Jan 20 at 4:47
LarryLarry
2,41331129
answered Jan 20 at 4:47
LarryLarry
2,41331129
answered Jan 20 at 4:47
LarryLarry
2,41331129
answered Jan 20 at 4:47
LarryLarry
2,41331129
2,41331129
$begingroup$
Thanks So Much! Please go through this once artofproblemsolving.com/community/q2h1766346p11567701 I just want to know how to decide which 'x' are safe to replace by their constant values?
$endgroup$
– Navneet Kumar
Jan 20 at 5:03
$begingroup$
Nikolas did a pretty good job explaining when it is safe to replace the values. Yes, when the denominator becomes zero, you don't want to replace the value. In this question, the denominator is zero, so you can't just replace the value. You would instead do L'Hospital or some other manipulations as presented in other answers posted here.
$endgroup$
– Larry
Jan 20 at 5:48
add a comment |
$begingroup$
Thanks So Much! Please go through this once artofproblemsolving.com/community/q2h1766346p11567701 I just want to know how to decide which 'x' are safe to replace by their constant values?
$endgroup$
– Navneet Kumar
Jan 20 at 5:03
$begingroup$
Nikolas did a pretty good job explaining when it is safe to replace the values. Yes, when the denominator becomes zero, you don't want to replace the value. In this question, the denominator is zero, so you can't just replace the value. You would instead do L'Hospital or some other manipulations as presented in other answers posted here.
$endgroup$
– Larry
Jan 20 at 5:48
$begingroup$
Thanks So Much! Please go through this once artofproblemsolving.com/community/q2h1766346p11567701 I just want to know how to decide which 'x' are safe to replace by their constant values?
$endgroup$
– Navneet Kumar
Jan 20 at 5:03
$begingroup$
Thanks So Much! Please go through this once artofproblemsolving.com/community/q2h1766346p11567701 I just want to know how to decide which 'x' are safe to replace by their constant values?
$endgroup$
– Navneet Kumar
Jan 20 at 5:03
$begingroup$
Nikolas did a pretty good job explaining when it is safe to replace the values. Yes, when the denominator becomes zero, you don't want to replace the value. In this question, the denominator is zero, so you can't just replace the value. You would instead do L'Hospital or some other manipulations as presented in other answers posted here.
$endgroup$
– Larry
Jan 20 at 5:48
$begingroup$
Nikolas did a pretty good job explaining when it is safe to replace the values. Yes, when the denominator becomes zero, you don't want to replace the value. In this question, the denominator is zero, so you can't just replace the value. You would instead do L'Hospital or some other manipulations as presented in other answers posted here.
$endgroup$
– Larry
Jan 20 at 5:48
add a comment |
$begingroup$
Hint
Compose Taylor series
$$cos(x)=1-frac{x^2}{2}+Oleft(x^4right)$$
$$cos left(frac{pi}{2} cos (x)right)=sin left(frac{pi }{4}x^2+Oleft(x^4right)right)$$ The next step is simple.
answered Jan 20 at 4:43
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
add a comment |
$begingroup$
Hint
Compose Taylor series
$$cos(x)=1-frac{x^2}{2}+Oleft(x^4right)$$
$$cos left(frac{pi}{2} cos (x)right)=sin left(frac{pi }{4}x^2+Oleft(x^4right)right)$$ The next step is simple.
answered Jan 20 at 4:43
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
add a comment |
$begingroup$
Hint
Compose Taylor series
$$cos(x)=1-frac{x^2}{2}+Oleft(x^4right)$$
$$cos left(frac{pi}{2} cos (x)right)=sin left(frac{pi }{4}x^2+Oleft(x^4right)right)$$ The next step is simple.
answered Jan 20 at 4:43
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
Hint
Compose Taylor series
$$cos(x)=1-frac{x^2}{2}+Oleft(x^4right)$$
$$cos left(frac{pi}{2} cos (x)right)=sin left(frac{pi }{4}x^2+Oleft(x^4right)right)$$ The next step is simple.
answered Jan 20 at 4:43
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 20 at 4:43
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 20 at 4:43
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 20 at 4:43
Claude LeiboviciClaude Leibovici
123k1157134
123k1157134
add a comment |
add a comment |
$begingroup$
Using $frac{1-cos(x)}{2}=sin^2(x/2)$ and $lim_{tto 0}frac{sin(t)}{t}=1$, we can write
$$begin{align}
frac{cosleft(fracpi2 cos(x)right)}{x^2}&=frac{sinleft(pi sin^2(x/2)right)}{x^2}\\
&=fracpi4underbrace{left(frac{sin(x/2)}{x/2}right)^2}_{to1,,text{as},,xto0}underbrace{left(frac{sinleft(pi sin^2(x/2)right)}{pisin^2(x/2)}right)}_{to 1,,text{as},,xto0}\\
end{align}$$
Therefore, we find that
$$lim_{xto0}frac{cosleft(fracpi2 cos(x)right)}{x^2}=fracpi4$$
answered Jan 20 at 5:19
Mark ViolaMark Viola
133k1277175
$endgroup$
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
$endgroup$
– Mark Viola
Jan 30 at 5:07
add a comment |
$begingroup$
Using $frac{1-cos(x)}{2}=sin^2(x/2)$ and $lim_{tto 0}frac{sin(t)}{t}=1$, we can write
$$begin{align}
frac{cosleft(fracpi2 cos(x)right)}{x^2}&=frac{sinleft(pi sin^2(x/2)right)}{x^2}\\
&=fracpi4underbrace{left(frac{sin(x/2)}{x/2}right)^2}_{to1,,text{as},,xto0}underbrace{left(frac{sinleft(pi sin^2(x/2)right)}{pisin^2(x/2)}right)}_{to 1,,text{as},,xto0}\\
end{align}$$
Therefore, we find that
$$lim_{xto0}frac{cosleft(fracpi2 cos(x)right)}{x^2}=fracpi4$$
answered Jan 20 at 5:19
Mark ViolaMark Viola
133k1277175
$endgroup$
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
$endgroup$
– Mark Viola
Jan 30 at 5:07
add a comment |
$begingroup$
Using $frac{1-cos(x)}{2}=sin^2(x/2)$ and $lim_{tto 0}frac{sin(t)}{t}=1$, we can write
$$begin{align}
frac{cosleft(fracpi2 cos(x)right)}{x^2}&=frac{sinleft(pi sin^2(x/2)right)}{x^2}\\
&=fracpi4underbrace{left(frac{sin(x/2)}{x/2}right)^2}_{to1,,text{as},,xto0}underbrace{left(frac{sinleft(pi sin^2(x/2)right)}{pisin^2(x/2)}right)}_{to 1,,text{as},,xto0}\\
end{align}$$
Therefore, we find that
$$lim_{xto0}frac{cosleft(fracpi2 cos(x)right)}{x^2}=fracpi4$$
answered Jan 20 at 5:19
Mark ViolaMark Viola
133k1277175
$endgroup$
Using $frac{1-cos(x)}{2}=sin^2(x/2)$ and $lim_{tto 0}frac{sin(t)}{t}=1$, we can write
$$begin{align}
frac{cosleft(fracpi2 cos(x)right)}{x^2}&=frac{sinleft(pi sin^2(x/2)right)}{x^2}\\
&=fracpi4underbrace{left(frac{sin(x/2)}{x/2}right)^2}_{to1,,text{as},,xto0}underbrace{left(frac{sinleft(pi sin^2(x/2)right)}{pisin^2(x/2)}right)}_{to 1,,text{as},,xto0}\\
end{align}$$
Therefore, we find that
$$lim_{xto0}frac{cosleft(fracpi2 cos(x)right)}{x^2}=fracpi4$$
answered Jan 20 at 5:19
Mark ViolaMark Viola
133k1277175
answered Jan 20 at 5:19
Mark ViolaMark Viola
133k1277175
answered Jan 20 at 5:19
Mark ViolaMark Viola
133k1277175
answered Jan 20 at 5:19
Mark ViolaMark Viola
133k1277175
133k1277175
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Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
$endgroup$
– Mark Viola
Jan 30 at 5:07
add a comment |
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
$endgroup$
– Mark Viola
Jan 30 at 5:07
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
$endgroup$
– Mark Viola
Jan 30 at 5:07
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
$endgroup$
– Mark Viola
Jan 30 at 5:07
add a comment |
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You can't simply plug 0 into parts. Take $limlimits_{xto0}frac xx$ for example. Plugging 0 into the numerator implies the limit should be 0, but clearly the limit is 1.
$endgroup$
– Simply Beautiful Art
Jan 20 at 4:49
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@SimplyBeautifulArt But,in some of the questions we do replace 'x' by the constant values such that it simplifies to a non zero number.Why are we allowed to do that?
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– Navneet Kumar
Jan 20 at 4:57
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Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
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– Simply Beautiful Art
Jan 20 at 4:58
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@SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
$endgroup$
– Navneet Kumar
Jan 20 at 5:01
1
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"The only thing that I need to keep in mind is that the expression doesn't become indeterminate." This is vague, and also the reason we have clear limit rules to follow.
$endgroup$
– Simply Beautiful Art
Jan 20 at 5:06