3
$\begingroup$

I am trying to solve the limit: $$\lim_{x\to\infty}x^\frac{5}{3}\left(\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-x^\frac{1}{3}\right)$$

I was trying to find a way to bring it into a fraction form to apply L'Hospital's rule, and I tried using $$a-b=\frac{a^3-b^3}{a^2+ab+b^2}$$

But it made it even more complex and after applying L'Hospital's rule I got stuck with all the terms. Is there a smarter way to evaluate it?

$\endgroup$
  • 1
    $\begingroup$ Can you use Taylor's theorem? $\endgroup$ – Peter Foreman Feb 25 at 17:29
  • $\begingroup$ Let L be the expression inside the limit. Then try taking logarithm on both sides and push the limit inside the logarithm. $\endgroup$ – gruana Feb 25 at 17:39
3
$\begingroup$

First I would factor $x^{1/3}$ out of the second term: $$ x^{5/3}\left(\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-x^\frac{1}{3}\right) = x^{2}\left(\left(1+ \frac{1}{x}\sin\left(\frac{1}{x}\right)\right)^{1/3}-1\right) $$ Then I would substitute $t = \frac{1}{x}$: $$ \lim_{x\to\infty}x^{2}\left(\left(1+ \frac{1}{x}\sin\left(\frac{1}{x}\right)\right)^{1/3}-1\right) = \lim_{t\to 0^+} \frac{(1+t \sin t)^{1/3} - 1}{t^2} $$ At this point we could use Taylor's Theorem, or the Binomial series, or L'Hôpital's rule. The limit of the quotient of the derivatives is: \begin{align*} \lim_{t\to 0^+} \frac{\frac{1}{3}(1+t\sin t)^{-2/3}(t \cos t + \sin t)}{2t} &= \frac{1}{6}\cdot 1 \cdot \lim_{t\to 0^+} \left(\cos t + \frac{\sin t}{t}\right) \\&= \frac{1}{6}(1+1) = \frac{1}{3} \end{align*}

$\endgroup$
2
$\begingroup$

$$\lim_{x\to\infty}x^\frac{5}{3}\left(\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-x^\frac{1}{3}\right)= \lim_{x\to\infty}x^2\left(\left(1+\frac{1}{x}\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-1\right)=\lim_{x\to\infty}\frac{\left(1+\frac{1}{x}\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-1}{\frac{1}{x^2}}$$

Now using L'Hopital we get

$$\lim_{x\to\infty}\frac{\frac{1}{3}\left(1+\frac{1}{x}\sin\left(\frac{1}{x}\right)\right)^\frac{-2}{3}\left(\frac{-1}{x^2}\right)\left( \sin\left(\frac{1}{x}\right) +\frac{1}{x}\cos\left(\frac{1}{x}\right) \right)}{\frac{-2}{x^3}}=\frac{1}{6}\lim_{x\to\infty}\frac{\sin\left(\frac{1}{x}\right) +\frac{1}{x}\cos\left(\frac{1}{x}\right)}{\frac{1}{x}}=\\ \frac{1}{6}\lim_{x\to\infty}\frac{\frac{-2}{x^2}\cos\left(\frac{1}{x}\right)-\frac{-1}{x^3}\sin\left(\frac{1}{x}\right)}{\frac{-1}{x^2}}=\frac{1}{6}\times2=\frac{1}{3}$$

$\endgroup$
0
$\begingroup$

I think using the difference of cubes should work:$$\lim_{x\to\infty}x^\frac{5}{3}\left(\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-x^\frac{1}{3}\right)=\\ \lim_{x\to\infty}\frac{x^\frac{5}{3}\sin\frac{1}{x}}{\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{2}{3}+\left(x+\sin\frac{1}{x}\right)^{\frac{1}{3}}x^\frac{1}{3}+x^\frac{2}{3}}=\\ \lim_{x\to\infty}\frac{x^\frac{5}{3}x^{-1}}{\left(x+x^{-1}\right)^\frac{2}{3}+\left(x+x^{-1}\right)^\frac{1}{3}x^\frac{1}{3}+x^\frac{2}{3}}=\\ \lim_{x\to\infty}\frac{x^\frac{2}{3}}{x^\frac{2}{3}+x^\frac{2}{3}+x^\frac{2}{3}}=\frac{1}{3} $$

$\endgroup$
  • $\begingroup$ Is it ok to just subsitute 1/x for sin(1/x)? I thought you can only do that when it can be seperated out as a sum or product of limits. $\endgroup$ – Noah Dorsey Feb 25 at 18:42
  • $\begingroup$ @NoahDorsey: yes, because in this case $\lim_{1/x \to 0} \frac{{\sin 1/x}}{1/x} = 1$ $\endgroup$ – Vasya Feb 25 at 18:50
  • $\begingroup$ I thought you couldn't do that if it is an indeterminate form, because then: $\lim_{x\to0}x=0$ so $\lim_{x\to0}\frac{x}{x}=\lim_{x\to0}\frac{0}{x}=0$ Which is wrong. $\endgroup$ – Noah Dorsey Feb 25 at 19:04
0
$\begingroup$

You have that

$$ \begin{gathered} \mathop {\lim }\limits_{x \to + \infty } x^{\frac{5} {3}} \left[ {\left( {x + \sin \left( {\frac{1} {x}} \right)} \right)^{\frac{1} {3}} - x^{\frac{1} {3}} } \right] = \hfill \\ \hfill \\ \mathop {\lim }\limits_{x \to + \infty } x^{\frac{5} {3}} \left[ {x^{\frac{1} {3}} \left( {1 + \frac{1} {x}\sin \left( {\frac{1} {x}} \right)} \right)^{\frac{1} {3}} - 1} \right] = \hfill \\ \hfill \\ = \mathop {\lim }\limits_{x \to + \infty } x^2 \left[ {\frac{1} {{3x}}\sin \left( {\frac{1} {x}} \right)} \right] = \hfill \\ \hfill \\ = \frac{1} {3}\mathop {\lim }\limits_{x \to + \infty } x\left[ {\sin \left( {\frac{1} {x}} \right)} \right] = \frac{1} {3}\mathop {\lim }\limits_{x \to + \infty } \left[ {\frac{{\sin \left( {\frac{1} {x}} \right)}} {{\frac{1} {x}}}} \right] = \frac{1} {3} \hfill \\ \end{gathered} $$ where we used the fact that $$ \left( {1 + f\left( x \right)} \right)^\alpha - 1 \sim \alpha f(x)\,\,\,\,\,\,\left( {x \to x_0 } \right) $$ provided $$ \mathop {\lim }\limits_{x \to x_0 } f(x) = 0 $$ and the fact that $$ \mathop {\lim }\limits_{t \to 0} \frac{{\sin t}} {t} = 1 $$ with $t=1/x$ as $x \to +\infty$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.