Mixing
Finding limit as $x rightarrow 0$
$begingroup$
The question asks to find
$$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$
I solved it by applying the L'Hopital rule since it is of the form $frac{0}{0}$ and then once I differentiated we get $$lim_{x rightarrow0} left(frac{frac{d sin x^n}{dx}}{frac{ left(d sin x right)^m}{dx}}right),
$$
which gives us
$$
lim_{x rightarrow0} frac{ncos x cdot x^{n-1}}{left(m sin xright)^{m-1} cos x}.
$$
Now this gives us a function of the form $$frac{n x^{n-1}}{left(m sin xright)^{m-1}},
$$
which means that since $n>m$ now, we can keep differentiating the denominator and we will eventually get something of the form $frac{0}{k}$ which means the limit must be $0$.
This is how I have gone about it. I have $2$ questions:
$1.$ Is there a better method than this to solve it?
$2.$ Can we find out the limit of this function if $m>n$?
calculus limits
edited Jan 26 at 8:35
Mee Seong Im
2,8151617
asked Jan 26 at 8:20
Prakhar NagpalPrakhar Nagpal
752318
$endgroup$
add a comment |
$begingroup$
The question asks to find
$$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$
I solved it by applying the L'Hopital rule since it is of the form $frac{0}{0}$ and then once I differentiated we get $$lim_{x rightarrow0} left(frac{frac{d sin x^n}{dx}}{frac{ left(d sin x right)^m}{dx}}right),
$$
which gives us
$$
lim_{x rightarrow0} frac{ncos x cdot x^{n-1}}{left(m sin xright)^{m-1} cos x}.
$$
Now this gives us a function of the form $$frac{n x^{n-1}}{left(m sin xright)^{m-1}},
$$
which means that since $n>m$ now, we can keep differentiating the denominator and we will eventually get something of the form $frac{0}{k}$ which means the limit must be $0$.
This is how I have gone about it. I have $2$ questions:
$1.$ Is there a better method than this to solve it?
$2.$ Can we find out the limit of this function if $m>n$?
calculus limits
edited Jan 26 at 8:35
Mee Seong Im
2,8151617
asked Jan 26 at 8:20
Prakhar NagpalPrakhar Nagpal
752318
$endgroup$
add a comment |
$begingroup$
The question asks to find
$$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$
I solved it by applying the L'Hopital rule since it is of the form $frac{0}{0}$ and then once I differentiated we get $$lim_{x rightarrow0} left(frac{frac{d sin x^n}{dx}}{frac{ left(d sin x right)^m}{dx}}right),
$$
which gives us
$$
lim_{x rightarrow0} frac{ncos x cdot x^{n-1}}{left(m sin xright)^{m-1} cos x}.
$$
Now this gives us a function of the form $$frac{n x^{n-1}}{left(m sin xright)^{m-1}},
$$
which means that since $n>m$ now, we can keep differentiating the denominator and we will eventually get something of the form $frac{0}{k}$ which means the limit must be $0$.
This is how I have gone about it. I have $2$ questions:
$1.$ Is there a better method than this to solve it?
$2.$ Can we find out the limit of this function if $m>n$?
calculus limits
edited Jan 26 at 8:35
Mee Seong Im
2,8151617
asked Jan 26 at 8:20
Prakhar NagpalPrakhar Nagpal
752318
$endgroup$
The question asks to find
$$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$
I solved it by applying the L'Hopital rule since it is of the form $frac{0}{0}$ and then once I differentiated we get $$lim_{x rightarrow0} left(frac{frac{d sin x^n}{dx}}{frac{ left(d sin x right)^m}{dx}}right),
$$
which gives us
$$
lim_{x rightarrow0} frac{ncos x cdot x^{n-1}}{left(m sin xright)^{m-1} cos x}.
$$
Now this gives us a function of the form $$frac{n x^{n-1}}{left(m sin xright)^{m-1}},
$$
which means that since $n>m$ now, we can keep differentiating the denominator and we will eventually get something of the form $frac{0}{k}$ which means the limit must be $0$.
This is how I have gone about it. I have $2$ questions:
$1.$ Is there a better method than this to solve it?
$2.$ Can we find out the limit of this function if $m>n$?
calculus limits
calculus limits
edited Jan 26 at 8:35
Mee Seong Im
2,8151617
asked Jan 26 at 8:20
Prakhar NagpalPrakhar Nagpal
752318
edited Jan 26 at 8:35
Mee Seong Im
2,8151617
asked Jan 26 at 8:20
Prakhar NagpalPrakhar Nagpal
752318
edited Jan 26 at 8:35
Mee Seong Im
2,8151617
edited Jan 26 at 8:35
Mee Seong Im
2,8151617
edited Jan 26 at 8:35
Mee Seong Im
2,8151617
2,8151617
asked Jan 26 at 8:20
Prakhar NagpalPrakhar Nagpal
752318
asked Jan 26 at 8:20
Prakhar NagpalPrakhar Nagpal
752318
asked Jan 26 at 8:20
Prakhar NagpalPrakhar Nagpal
752318
752318
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Write the function as follows:
$$ frac{ sin x^n }{ x^n } cdot frac{x^m}{sin^m x} cdot x^{n-m}$$
Now, in the limit we obtain $1 cdot 1^m cdot 0 = 0 $ since $n-m>0$. We have used the elementary fact that
$$ lim_{x to 0} frac{ sin x }{x} = 1 $$
answered Jan 26 at 8:24
Jimmy SabaterJimmy Sabater
3,054325
$endgroup$
add a comment |
$begingroup$
$$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$
$$= lim limits_{x rightarrow 0} frac{x^n (1 - frac{x^{2n}}{3!}dots)}{x^m[ 1 - frac{x^2}{3!} dots]^m} = limlimits_{x rightarrow0} x^{n-m} =0$$
By the way, $sin x = x = frac {x^3}{3!} + frac{x^5}{5!} dots$
and $cos x = 1 - frac {x^2}{2!} + frac{x^4}{4!} dots$
answered Jan 27 at 12:45
Abhas Kumar SinhaAbhas Kumar Sinha
304115
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Write the function as follows:
$$ frac{ sin x^n }{ x^n } cdot frac{x^m}{sin^m x} cdot x^{n-m}$$
Now, in the limit we obtain $1 cdot 1^m cdot 0 = 0 $ since $n-m>0$. We have used the elementary fact that
$$ lim_{x to 0} frac{ sin x }{x} = 1 $$
answered Jan 26 at 8:24
Jimmy SabaterJimmy Sabater
3,054325
$endgroup$
add a comment |
$begingroup$
Write the function as follows:
$$ frac{ sin x^n }{ x^n } cdot frac{x^m}{sin^m x} cdot x^{n-m}$$
Now, in the limit we obtain $1 cdot 1^m cdot 0 = 0 $ since $n-m>0$. We have used the elementary fact that
$$ lim_{x to 0} frac{ sin x }{x} = 1 $$
answered Jan 26 at 8:24
Jimmy SabaterJimmy Sabater
3,054325
$endgroup$
add a comment |
$begingroup$
Write the function as follows:
$$ frac{ sin x^n }{ x^n } cdot frac{x^m}{sin^m x} cdot x^{n-m}$$
Now, in the limit we obtain $1 cdot 1^m cdot 0 = 0 $ since $n-m>0$. We have used the elementary fact that
$$ lim_{x to 0} frac{ sin x }{x} = 1 $$
answered Jan 26 at 8:24
Jimmy SabaterJimmy Sabater
3,054325
$endgroup$
Write the function as follows:
$$ frac{ sin x^n }{ x^n } cdot frac{x^m}{sin^m x} cdot x^{n-m}$$
Now, in the limit we obtain $1 cdot 1^m cdot 0 = 0 $ since $n-m>0$. We have used the elementary fact that
$$ lim_{x to 0} frac{ sin x }{x} = 1 $$
answered Jan 26 at 8:24
Jimmy SabaterJimmy Sabater
3,054325
answered Jan 26 at 8:24
Jimmy SabaterJimmy Sabater
3,054325
answered Jan 26 at 8:24
Jimmy SabaterJimmy Sabater
3,054325
answered Jan 26 at 8:24
Jimmy SabaterJimmy Sabater
3,054325
3,054325
add a comment |
add a comment |
$begingroup$
$$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$
$$= lim limits_{x rightarrow 0} frac{x^n (1 - frac{x^{2n}}{3!}dots)}{x^m[ 1 - frac{x^2}{3!} dots]^m} = limlimits_{x rightarrow0} x^{n-m} =0$$
By the way, $sin x = x = frac {x^3}{3!} + frac{x^5}{5!} dots$
and $cos x = 1 - frac {x^2}{2!} + frac{x^4}{4!} dots$
answered Jan 27 at 12:45
Abhas Kumar SinhaAbhas Kumar Sinha
304115
$endgroup$
add a comment |
$begingroup$
$$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$
$$= lim limits_{x rightarrow 0} frac{x^n (1 - frac{x^{2n}}{3!}dots)}{x^m[ 1 - frac{x^2}{3!} dots]^m} = limlimits_{x rightarrow0} x^{n-m} =0$$
By the way, $sin x = x = frac {x^3}{3!} + frac{x^5}{5!} dots$
and $cos x = 1 - frac {x^2}{2!} + frac{x^4}{4!} dots$
answered Jan 27 at 12:45
Abhas Kumar SinhaAbhas Kumar Sinha
304115
$endgroup$
add a comment |
$begingroup$
$$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$
$$= lim limits_{x rightarrow 0} frac{x^n (1 - frac{x^{2n}}{3!}dots)}{x^m[ 1 - frac{x^2}{3!} dots]^m} = limlimits_{x rightarrow0} x^{n-m} =0$$
By the way, $sin x = x = frac {x^3}{3!} + frac{x^5}{5!} dots$
and $cos x = 1 - frac {x^2}{2!} + frac{x^4}{4!} dots$
answered Jan 27 at 12:45
Abhas Kumar SinhaAbhas Kumar Sinha
304115
$endgroup$
$$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$
$$= lim limits_{x rightarrow 0} frac{x^n (1 - frac{x^{2n}}{3!}dots)}{x^m[ 1 - frac{x^2}{3!} dots]^m} = limlimits_{x rightarrow0} x^{n-m} =0$$
By the way, $sin x = x = frac {x^3}{3!} + frac{x^5}{5!} dots$
and $cos x = 1 - frac {x^2}{2!} + frac{x^4}{4!} dots$
answered Jan 27 at 12:45
Abhas Kumar SinhaAbhas Kumar Sinha
304115
edited Jan 27 at 12:51
answered Jan 27 at 12:45
Abhas Kumar SinhaAbhas Kumar Sinha
304115
answered Jan 27 at 12:45
Abhas Kumar SinhaAbhas Kumar Sinha
304115
answered Jan 27 at 12:45
Abhas Kumar SinhaAbhas Kumar Sinha
304115
304115
add a comment |
add a comment |
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