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I try to find $\lim_{x \to \infty} x^{(2-\sin(\frac{2}{x}))}(x\sin(\frac{2}{x})-2)$ with L'Hospital's Rule but get stuck.

Here is my attempt. Let substitute $y = \dfrac{1}{x}$

\begin{align*} \lim_{x \to \infty} x^{(2-\sin(\frac{2}{x}))}(x\sin(\frac{2}{x})-2) &= \lim_{y \to 0} \frac{\left(\frac{\sin(2y)}{y}-2\right)}{y^{(2-\sin(2y))}} \\ &= \lim_{y \to 0} \frac{\sin(2y)-2y}{y^{(3-\sin(2y))}} \\ \text{(L'Hospital's Rule)} &= \lim_{y \to 0} \frac{2\cos(2y)-2}{-y^{(2-\sin(2y))}(\sin(2y) + 2y\ln(y)cos(2y) - 3)} \end{align*}

After I use L'Hospital's Rule, everything seem to get messier.

I think the next step is using L'Hospital's Rule again but everything will only get messier.

I think I go in the wrong way since L'Hospital's Rule should not make everything worse.

Where do I do wrong ?

user
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User1568736889
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2 Answers2

2

Notice that you can rewrite the first expression as $\frac{x^{2}(x\sin(2/x)-2)}{x^{\sin(2/x)}}$, then you can prove that $\lim_{x\to +\infty}x^{2}(x\sin(2/x)-2)=-4/3$ and using L'Hôpital's rule that $\lim_{x\to +\infty}x^{\sin(2/x)}=\lim_{x\to +\infty}\exp(\ln(x^{\sin(2/x)}))=1$.

A. P.
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We have that

$$\frac{\sin(2y)-2y}{y^{(3-\sin(2y))}}=\frac{\sin(2y)-2y}{y^3}\frac{y^3}{y^{(3-\sin(2y))}}=\frac{\sin(2y)-2y}{y^3}\frac{1}{y^{(-\sin(2y))}}$$

then observe

$$\lim_{y \to 0}y^{(-\sin(2y))}=\lim_{y \to 0}\left(y^y\right)^{\frac{-\sin(2y)}{y}}$$

which is solveable by standard limits and apply l'Hospital to $\lim_{y \to 0}\frac{\sin(2y)-2y}{y^3}$.

user
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