Evaluating limits

Question

Evaluate the limit without L’Hôpital rule: $$ \lim_{x \to 0}\frac{\sin^2{x}+2\ln\left(\cos{x}\right)}{x^4} $$

My work is: \begin{align} L&=\lim_{x \to 0}\frac{\sin^2{x}-x^2}{x^4}+\lim_{x \to 0} \frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\ &= \lim_{x \to 0}\frac{\sin{x}-x}{x^3} \lim_{x \to 0}\frac{\sin{x}+x}{x}+ \lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\ &= \frac{-1}{6}\left[\lim_{x \to 0}\frac{\sin x}{x}+1\right] +\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\ &=\frac{-1}{6}\left(2\right)+\lim \limits_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\ &= \frac{-1}{3}+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4} \end{align} I could not evaluate the second limit

Answer 1

I don't think there is a way to evaluate the limit without using the l'Hospital rule in some form. (Using Taylor series is basically the same, because they are obtained by calculating the derivatives.)

I would calculate the limit by applying l'Hospital once and then using known series expansions: \begin{align} &~~~~\lim_{x \to 0}\frac{\sin^2{x}+2\ln\left(\cos{x}\right)}{x^4} \\ &=\lim_{x \to 0} \frac{ \sin(2x)- 2 \tan(x)}{4x^3} \\\ &=\lim_{x \to 0} \frac{2x-\frac 8 6 x^3 + \mathcal O(x^5) - 2(x + \frac 1 3 x^3 +\mathcal O(x^5)) } {4 x^3} \\ &= \lim_{x \to 0}\frac{-2 x^3 + \mathcal O(x^5)}{4x^3} = -\frac 1 2 \end{align}

Answer 2

With Taylor expansion

$$\sin (x)=x-\frac {x^3}{6}+x^4\epsilon (x) $$

$$\sin^2 (x)=x^2-\frac {x^4}{3}+x^5\epsilon (x) $$

$$\cos (x)-1=-\frac {x^2}{2}+\frac {x^4}{24}+x^5\epsilon (x) $$ $$\ln (X+1)=X-\frac {X^2}{2}+X^2\epsilon (X) $$

$$\ln (\cos (x))=\ln \Bigl(\cos (x)-1+1\Bigr) $$ $$=-\frac {x^2}{2}+\frac {x^4}{24}-\frac {x^4}{8}+x^5\epsilon (x) $$

Replacing that, you should find

$$L=-\frac {1}{3}-\frac {1}{4}+\frac {1}{12}=-\frac {1}{2} $$

Answer 3

Or we could've approached it differently from the beginning, if we observe that $$2\ln(\cos x)=\ln(\cos^2 x)=\ln(1-\sin^2 x)$$ and use the series for $$\ln(1-t)=-\sum_{n=1}^{\infty}\frac{t^n}{n}=-t-\frac{t^2}{2}-\frac{t^3}{3}-\cdots.$$

Then for the original limit: $$\lim_{x\to0}\frac{\sin^2 x+2\ln(\cos x)}{x^4}= \lim_{x\to0}\frac{\sin^2 x+\ln(1-\sin^2 x)}{x^4}= \lim_{x\to0}\frac{\sin^2 x-\sin^2 x-\frac{\sin^4 x}{2}-O(\sin^6 x)}{x^4}= \lim_{x\to0}\frac{-\frac{\sin^4 x}{2}-O(x^6)}{x^4}= -\frac{1}{2}.$$

Answer 4

Your last line is $$L=-\frac{1}{3}+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}$$ So, as other answers used, by Taylor around $x=0$, $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ $$\log(\cos(x))=-\frac{x^2}{2}-\frac{x^4}{12}+O\left(x^6\right)$$ $$x^2+2\ln\left(\cos{x}\right)=-\frac{x^4}{6}+O\left(x^6\right)$$ $$\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}=-\frac{1}{6}+O\left(x^2\right)$$ making $$L=-\frac{1}{3}-\frac{1}{6}=-\frac{1}{2}$$

Answer 5

We evaluate the limit in question as follows \begin{align} L&=\lim_{x\to 0}\frac{\sin^{2}x+2\log\cos x}{x^{4}}\notag\\ &=\lim_{x\to 0} \frac{\log(1-\sin^{2}) +\sin^{2}x}{\sin^{4}x}\cdot\frac{\sin^{4}x}{x^{4}}\notag\\ &=\lim_{t\to 0^{+}}\frac{\log(1-t)+t}{t^{2}}\text{ (putting } t=\sin^{2}x)\notag\\ &=-\frac{1}{2}\text{ (via Taylor series or L'Hospital's Rule)} \notag \end{align} I have left the last step (which is almost routine).

Answer 6

This is my solution after alot of tries Note:The special limit which used in the solution can be proved without niether Lospital rule nor series