Solving a limit by the Squeeze theorem

I want to compute $$\lim_{x\rightarrow 0} \left(\frac{x}{x^2 + \sin x} \right).$$

By L'Hopital's rule, we can simply differentiate the numerator and the denominator with respect to $$x$$ to obtain

\begin{align*} \lim_{x \rightarrow 0} \left(\frac{1}{2x + \cos x} \right) = 1. \end{align*}

My question: I want to use the squeeze theorem to evaluate the above limit. Is it possible?

We know that $$-1 \leq \sin x \leq 1$$, so we have:

\begin{align*} \frac{x}{x^2 -1}\leq \frac{x}{x^2 + \sin x} \leq \frac{x}{x^2 +1} \end{align*}

But here I obtain a limit between $$-1$$ and $$1$$. I know that I am not using the squeeze theorem correctly. Is it possible to use it? If yes, how?

• Off point : alternative : avoid L'Hopital's rule and avoid squeeze theorem. If $~\lim_{x \to 0} f(x) ~$ exists and is not equal to $~0,~$ then you can routinely calculate $~\lim_{x\to 0} \dfrac{1}{f(x)}.~$ So, examine the reciprocal, utilizing the Taylor series. Nov 8, 2023 at 10:08
• "$-1 \leq \sin x \leq 1$ so we have $\frac{x}{x^2 -1}\leq \frac{x}{x^2 + \sin x} \leq \frac{x}{x^2 +1}$" is false. Nov 8, 2023 at 10:20
• OP - What is your definition of $\sin x?$ Nov 8, 2023 at 12:42

$$\frac x{x^2 + \sin x}=\frac1{x+ \frac{\sin x} x},$$ so your problem is essentially to prove that $$\lim_{x\to0}\frac{\sin x}x=1.$$ There are various ways to prove it but in any case, $$-1\le\sin x\le1$$ does not provide sufficient knowledge about the function $$\sin.$$ Even $$|\sin x|\le|x|$$ is not sufficient, since the zero function also satisfies it.

You can apply the squeeze theorem to $$|x\cos x|\le |\sin x|\le |x|.$$

Using Squeeze Theorem:
If you take into account that $$\sin(x) < x$$ for a small $$x$$ (this is, $$x \rightarrow 0$$), $$x^2 + \sin x < x^2 + x \Rightarrow$$: $$\frac{x}{x^2+\sin x} \geq \frac{x}{x^2+x} = \frac{x}{x(x+1)} = \frac{1}{x+1}$$whose limit is $$1$$, and it is a lower bound, and therefore, using Squeeze Theorem, the limit is $$1$$.
Note that $$x$$ could be either positive or negative, but this isn't a problem since in general $$|\sin x| < |x|$$ when $$x \rightarrow 0$$. Note that it is a lower bound, but your upper bound is incorrect. I have given you the lower one.

• Note that for $x \to 0$, you may have $x$ small and negative. It does not affect the result but you may need to adjust the argument. Nov 8, 2023 at 10:25
• And your lower bound is not sufficient. You need an upper bound. Nov 8, 2023 at 10:27
• You are right, thanks for the acclaration Nov 8, 2023 at 10:27
• The upper bound has been already offered by the poster Nov 8, 2023 at 10:28
• The poster's upper bound was wrong, as commented. Nov 8, 2023 at 10:31