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?
Best Answer
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*}