Calculating $\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^2\ln(\cos x)}$ without L'Hospital's rule

Question

this is my first post here so excuse the lack of knowledge about how things usually go.

My question revolves around calculating the limit as $x$ approaches $0$ of the following function:

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

The question came up in a test about a month ago and while I couldn't solve it in the test I've been working on it since then but I can't seem to get it. I know the limit is supposed to be $\frac{-2}{3}$ from some online calculators which abused l'hopital rule over and over again. I've tried playing around with it in so many ways but I always seem to get 0 over 0 or the so called indeterminate form. I've even tried calculating it by substituting in the Taylor series for the functions given but no luck. If anyone could show me a method of calculating this without using l'hopital rule or better yet, give me a hint as to how I should proceed I would be grateful.

Answer 1

More elementary solution (without Taylor series expansion or L'Hospital's rule) can be:

$$ \begin{align*} \lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^2\ln(\cos x)}&=\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^4}\cdot\frac{x^2}{1-\cos x}\cdot\frac{-(\cos x-1)}{\ln(\cos x)-0}\\ &=\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^4}\cdot\frac{x^2(1+\cos x)}{1-\cos^2x}\cdot\frac{-1}{\ln'(1)}\\ &=\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^4}\cdot\frac{2x^2}{\sin^2x}\cdot(-1)\\ &=-2\cdot\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^4}=L \end{align*} $$$$ \begin{align*} L_1&=\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^4}\\ &=\lim_{2x \to 0} \frac{\sin(4x^2)-\sin^2(2x)}{16x^4}=\lim_{x \to 0} \frac{2\sin(2x^2)\cos(2x^2)-4\sin^2(x)\cos^2(x)}{16x^4}\\ &=\lim_{x \to 0} \frac{4\sin(x^2)\cos(x^2)(1-2\sin^2(x^2))-4\sin^2(x)(1-\sin^2(x))}{16x^4}\\ &=\lim_{x \to 0} \frac{4\sin(x^2)(1-2\sin^2(x^2))-4\sin^2(x)(1-\sin^2(x))}{16x^4}\\ &=\lim_{x \to 0} \frac{4\sin(x^2)-4\sin^2(x)}{16x^4}-\lim_{x \to 0}\frac{8\sin^3(x^2)}{16x^4} +\lim_{x \to 0}\frac{4\sin^4(x)}{16x^4} \\ &=\frac{L_1}{4}-\lim_{x^2 \to 0}\frac{8}{16}\left(\frac{\sin x}{x}\right)^2\cdot\sin(x)+\lim_{x \to 0}\frac{4}{16}\left(\frac{\sin x}{x} \right)^4\\ &=\frac{L_1}{4}+\frac14\\ \therefore L_1&=\frac13 \end{align*} $$ $$\therefore L=-2L_1=-\frac23$$

Answer 2

HINT

Use the following Maclaurin series: $$\begin{align} \sin x&=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\\ \ln(1-x)&=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots\end{align}$$ along with the trig identity $\cos x\equiv1-2\sin^2\frac{x}{2}$.

Answer 3

You can always "cheat" about the use of Taylor's if you want to:

For example, since $\frac{1-\cos x}{x^2} \to \frac{1}{2}$, you can manipulate things so that $\frac{\cos x}{1-\frac{1}{2}x^2} \to 1$ and you can use some identities to prove that $\dfrac{\ln \cos x}{\ln \left( 1-\frac{1}{2}x^2 \right)} \to 1$. Additionally, you know that $\dfrac{\ln(1+y)}{y} \to 1$, so, again, you can hide everything and plug something like $$\dfrac{\sin (x^2)-\sin(x)^2}{x\left(-\frac{1}{2}x^2 \right)}\cdot \dfrac{-\frac{1}{2}x^2}{\ln(1-\frac{1}{2}x^2)}\cdot \dfrac{\ln(1-\frac{1}{2}x^2)}{\ln \cos x}$$

(you can continue with this idea as much as you want until you came up with "known" limits)

Answer 4

Using composition of Taylor series around $x=0$ $$ y=\frac{\sin(x^2)-\sin^2(x)}{x^2\log(\cos (x))}$$

$$\sin(x^2)=x^2-\frac{x^6}{6}+O\left(x^{10}\right)$$ $$\sin^2(x)=x^2-\frac{x^4}{3}+\frac{2 x^6}{45}+O\left(x^8\right)$$ $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ $$\log(\cos (x))=-\frac{x^2}{2}-\frac{x^4}{12}+O\left(x^6\right)$$ $$y=\frac{\frac{x^4}{3}-\frac{19 x^6}{90}+O\left(x^8\right) } {-\frac{x^4}{2}-\frac{x^6}{12}+O\left(x^8\right) }=-\frac{2}{3}+\frac{8 x^2}{15}+O\left(x^4\right)$$ shows the limit and how it is approached.