# Compute the limit without L'Hospital's rule

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

• Not quite. You need to write $\left| \frac{x^2\cos(x)}{\sin(x)}\right|\le \frac{x^2}{|\sin(x)|}=|x|\,\left|\frac{x}{\sin(x)}\right|\to 0\times 1$ – Mark Viola Jan 16 '18 at 19:23
• you should correct x tends to 0 not to $\infty$ – 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 $\sin x \to0.$

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

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

Then

\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 $$\smash{\dfrac{x^2\cos\dfrac1x}{\sin x}} = O(x)\to 0.$$