Compute $\lim\limits_{x\rightarrow 0}\frac{\cos{x}-1}{\ln{(1+\sin^2{x})}}$

Question

The limit is $$\lim\limits_{x\rightarrow 0}\frac{\cos{x}-1}{\ln{(1+\sin^2{x})}},$$

I'm trying to rewrite the function as $$\frac{\cos^2{x}}{\ln{(1+\sin^2{x})}}\cdot\frac{1}{\cos{x}}-\frac{1}{\ln{(1+\sin^2{x})}}$$

But I'm not sure how to proceed from here.

Note: that L'Hopital's rule is not allowed and I should only be able using sin, cos (no sec and csc).

EDIT:

I forgott to square the $\sin^2(x)$. Sorry about that :/

Answer 1

Taken from my comment:

Notice that

$$\frac{\cos(x)-1}{\ln(1+\sin^2(x))}=\frac{\cos(x)-1}{x^2}\left(\frac{\sin^2(x)}{x^2}\frac{\ln(1+u)}u\right)^{-1}$$

where $u=\sin^2(x)$. As $x\to0$, $u\to0$, and since we have, by definition of the derivative:

$$\frac d{dx}\sin(x)\bigg|_{x=0}=\lim_{x\to0}\frac{\sin(x)}x=1$$

$$\frac d{du}\ln(1+u)\bigg|_{u=0}=\lim_{u\to0}\frac{\ln(1+u)}u=1$$

and

$$\lim_{x\to0}\frac{\cos(x)-1}{x^2}=-\frac12$$

(which are all fundamental and well-known limits, the last may be derived via series or integral forms)

Thus,

$$\lim_{x\to0}\frac{\cos(x)-1}{\ln(1+\sin^2(x))}=-\frac12\cdot(1^2\cdot1)^{-1}=-\frac12$$

Answer 2

$$\frac{\cos x-1}{\log(1+\sin^2 x)} = \underbrace{\frac{-2\sin^2\frac{x}{2}}{\sin^2 x}}_{\to -1/2}\cdot\underbrace{\frac{\sin^2 x}{\log(1+\sin^2 x)}}_{\to 1}.$$

Answer 3

Using the series forms: $$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dotsb, \;\forall x$$ and $$\ln(1+x) = x - \frac{x^2}{2}+\frac{x^3}{3} - \dotsb,\;x\in(-1,1] $$ then \begin{align}\lim_{x\to 0} \frac{\cos(x)-1}{\ln(1+\sin^2(x))} &= \lim_{x\to 0} \frac{\left[1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dotsb\right] - 1}{\sin^2(x) - \frac{[\sin^2(x)]^2}{2}+\frac{[\sin^2(x)]^3}{3} - \dotsb}\\ &= \lim_{x\to 0} \frac{-\frac{x^2}{2!} + \frac{x^4}{4!} - \dotsb}{\sin^2(x)\left[1- \frac{\sin^2(x)}{2} + \frac{\sin^4(x)}{3} - \dotsb\right]} \cdot \frac{\tfrac{1}{x^2}}{\tfrac{1}{x^2}}\\ &= \lim_{x\to 0} \frac{-\frac12 +\frac{x^2}{4}-\dotsb}{\underbrace{\left(\frac{\sin(x)}{x}\right)^2}_{\to 1} \left[1- \underbrace{\frac{\sin^2(x)}{2} + \frac{\sin^4(x)}{3} - \dotsb}_{\to 0}\right] } = \frac{-\tfrac{1}{2} + 0}{1\cdot 1} = -\frac12 \end{align}