I'm not sure how to handle the trig functions with different arguments when computing this limit using L'Hospital's rule.

$$\lim_{x \rightarrow 0} \frac {x^2\cos(\frac {1} {x})} {\sin(x)}.$$

I have come up with the correct numerical answer via a different method, but am unsure if the logic would hold true for all cases (maybe I arrived at the correct answer by chance).

Here is my working:

Let $g(x)=x^2\cos(\frac 1 x)$ and $h(x) = \frac 1 {\sin x} = \csc x$.

We know that the following holds true for all $x$: $$-1 \le \cos (\frac 1 x) \le 1$$ Since $x^2 \ge 0$ for all x: $$-x^2 \le x^2\cos (\frac 1 x) \le x^2$$ Taking limits as $x \rightarrow \infty$ gives: $$ \lim_{x\rightarrow 0}(-x^2)\le \lim_{x\rightarrow 0}(x^2\cos (\frac 1 x)) \le \lim_{x\rightarrow 0}(x^2)$$ By the sandwich rule (or squeeze theorem): $$ 0 \le \lim_{x\rightarrow 0}(x^2\cos (\frac 1 x)) \le 0$$ $$ \Rightarrow \lim_{x\rightarrow 0}(x^2\cos (\frac 1 x)) $$ And hence, due to the algebra of limits: $$\lim_{x \rightarrow 0} \frac {x^2\cos(\frac {1} {x})} {\sin(x)} = 0.$$

  • 3
    $\begingroup$ Not quite. You need to write $\left| \frac{x^2\cos(x)}{\sin(x)}\right|\le \frac{x^2}{|\sin(x)|}=|x|\,\left|\frac{x}{\sin(x)}\right|\to 0\times 1$ $\endgroup$ – Mark Viola Jan 16 '18 at 19:23
  • $\begingroup$ you should correct x tends to 0 not to $\infty$ $\endgroup$ – Isham Jan 16 '18 at 19:26

$$ \frac {x^2\cos\frac 1 x} {\sin x} = x\cdot \frac x {\sin x} \cdot \cos\frac 1 x $$ Now use the fact that $-1 \le \cos \frac 1 x \le 1$ and $x\to0$ and one further fact not mentioned in your question: $$ \frac x {\sin x} \to 1 \text{ as } x\to0. $$ Without that last fact or something else other than what's in your question, you haven't dealt with the fact that $\sin x \to0.$

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    $\begingroup$ @Isham : So what are you saying needs correction? $\endgroup$ – Michael Hardy Jan 16 '18 at 19:26
  • $\begingroup$ +1 Sorry Michael it was for Op he wrote x tends to $ \infty$ $\endgroup$ – Isham Jan 16 '18 at 19:26

Note that $$\lim_{x \to 0}\frac{\sin(x)}{x} = \lim_{x \to 0}\frac{x}{\sin(x)} = 1$$


\begin{align} L=\lim_{x \to 0} \frac{x^2\cos(1/x)}{\sin(x)} &= \lim_{x \to 0}\frac{x}{\sin(x)}\frac{x \cos(1/x)}{1}\\ &=\lim_{x \to 0}x \cos(1/x) \end{align} And by Squeeze Theorem

$$-1 \le \cos(1/x) \le1$$ $$-x \le x\cos(1/x) \le x$$ $$0 \le \lim_{x \to 0}x \cos(1/x) \le 0$$ $$L=0$$


$$\begin{align}L&=\lim_\limits{x\to0}\dfrac x{\sin x}\cdot\lim_\limits{x\to0}x\cos\left(\dfrac1x\right)\\&=\lim_\limits{x\to0}x\cos\left(\dfrac1x\right)\end{align}$$

We know $$\begin{align}-1&\le\cos\left(\dfrac1x\right)\le1\\-x&\le x\cos\left(\dfrac1x\right)\le x\\\lim_\limits{x\to0}(-x)&\le\lim_\limits{x\to0} x\cos\left(\dfrac1x\right)\le\lim_\limits{x\to0}x\end{align}$$

So, $$\lim_\limits{x\to0} x\cos\left(\dfrac1x\right)=0$$ via the Squeeze Theorem

And hence $$L=0$$ via $$\lim_{n\to c} a_nb_n=\lim_\limits{n\to c}a_n\cdot\lim_\limits{n\to c}b_n$$


Use asymptotic analysis: you know $\sin x \sim_0 x$, hence $\dfrac{x^2}{\sin x} \sim_0 \dfrac{x^2}x=x $.

Furthermore, $\Bigl\vert\cos\dfrac1x \Bigr\vert \le 1$, so $$\smash{\dfrac{x^2\cos\dfrac1x}{\sin x}} = O(x)\to 0.$$


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