Evaluating $\lim\limits_{x \to 0} \frac{\cos\left(\sin^2 x\right)-1}{\sin x}$ without l'Hôpital

Question

I'm stuck trying to evaluate this limit without l'Hôpital's rule:

$$ \lim_{x \to 0} \frac{\cos\left(\sin^2x\right)-1}{\sin x} $$

Could anyone give me a hint on what I need to do? I tried many trig identities but I can't seem to find one that doesn't give me an indeterminate form.

Answer 1

Using $$ \lim_{x\to 0}\frac{1-\cos x}{x^2}=\lim_{x\to 0}\frac{2\sin^2(\frac {x}{2})}{x^2}=\frac12$$ one has $$\lim_{x \to 0} \frac{\cos\left(\sin^2x\right)-1}{\sin x}=-\lim_{x \to 0} \frac{1-\cos\left(\sin^2x\right)}{\sin x}=-\lim_{x \to 0} \frac{1-\cos\left(\sin^2x\right)}{\sin^2x}\frac{\sin^2x}{\sin x}=0.$$

Answer 2

We use:

i) $\lim\limits_{t \rightarrow 0} \frac{\sin t}{t}=1$,

ii) $\cos t - 1 = -2\sin^2\frac{t}{2}$.

Then

$\lim\limits_{x \rightarrow 0} \dfrac{\cos(\sin^2 x)-1}{\sin x}= \lim\limits_{x \rightarrow 0} \dfrac{-2\sin^2(\frac{\sin^2 x}{2})}{\sin x} = -\dfrac{1}{2}\lim\limits_{x \rightarrow 0} \dfrac{\sin^2(\frac{\sin^2 x}{2})}{(\frac{\sin^2 x}{2})^2}\sin^3 x = -\dfrac{1}{2}.(1)^2.0^3=0.$

Answer 3

My attempt, using asymptotic estimates. As $u \to 0$, $$ \cos u = 1 -\frac12 u^2 + O(u^4) . $$

As $x \to 0$, also $\sin x \to 0$, so \begin{align} \sin x &= x + O(x^3) \\ \frac{1}{\sin x} &= \frac{1}{x}+O(x) \\ \sin^2 x &= x^2 + O(x^4) \\ \cos(\sin^2 x) &= 1 -\frac12 \sin^2 x + O(\sin^4 x) = 1 -\frac12 (x^2 + O(x^4)) + O(x^4) = 1 -\frac12 x^2 + O(x^4) \\ \cos(\sin^2 x) - 1 &= -\frac12 x^2 + O(x^4) \\ \frac{\cos(\sin^2 x) - 1}{\sin x} &= \big(\frac{1}{x}+O(x)\big)\;\big(-\frac12 x^2 + O(x^4)\big) = -\frac12 x + O(x^3) \end{align} so $$ \lim_{x\to 0} \frac{\cos(\sin^2 x) - 1}{\sin x} = 0 . $$

Answer 4

Using derivatives:

$$\begin{align*} \lim_{x \to 0} \frac{\cos\left(\sin^2x\right)-1}{\sin x} &= \lim_{x \to 0} \frac{\cos\left(\sin^2x\right)-1}{\sin^2(x)} \cdot \lim_{x\to0} \frac{\sin(x)}x \cdot \lim_{x\to0} x \\[1ex] &= \frac{d\cos(y)}{dy}\bigg|_{y=0} \cdot \frac{d\sin(x)}{dx}\bigg|_{x=0} \cdot 0 & y=\sin^2(x) \\[1ex] &= -\sin(0) \cdot \cos(0) \cdot 0 \\[1ex] &= \boxed{0} \end{align*}$$