Evaluate the limit $\lim_{x \to 0} \left(\frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$

Question

Evaluate the limit $$\lim_{x \to 0}\left( \frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$$

My attempt

So we have $$\frac{1}{x^{2}}-\frac{\cos^{2}x}{\sin^{2}x}$$

$$=\frac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}$$ $$=\frac{x^2}{\sin^2 x}\cdot\frac{\sin x+x\cos x}{x}\cdot\frac{\sin x-x\cos x}{x^3}$$

Then I have $3$ limits to evaluate $$\lim_{x \to 0}\frac{x^2}{\sin^2 x}=\left(\lim_{x \to o}\frac{x}{\sin x}\right)^2=1^2=1$$

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

Now I'm having trouble with the last one which is $$\lim_{x \to 0}\frac{\sin x-x\cos x}{x^3}=?$$

Thanks for any help.

Answer 1

Using L'Hospital's rule (since direct evaluation gives $\bigl(\frac{0}{0}\bigr)$ ), we have the following:

$$\lim_{x \to 0} \frac{\cos x-\cos x +x\sin x}{3x^2}= \lim_{x \to 0} \frac{\sin x}{3x}.$$

We take the derivative of the numerator and denominator again:

$$\lim_{x \to 0} \frac{\cos x}{3} = \frac{1}{3}.$$

Answer 2

For your final problem I'd use L'Hospital's Rule to obtain:

$$\lim_{x \to 0} \frac{x \sin(x)}{3x^2} \implies \lim_{x \to 0} \frac{\sin(x)}{3x} \\ \hspace{.1cm} \text{using L'Hospital's again}, \hspace{.1cm} \\ \lim_{x \to 0}\frac{\cos(x)}{3} = \frac{1}{3}.$$

Answer 3

It is done much simpler in the following manner with just one application of LHR. \begin{align} L &= \lim_{x \to 0}\left(\frac{1}{x^{2}} - \frac{1}{\tan^{2}x}\right)\notag\\ &= \lim_{x \to 0}\frac{\tan^{2}x - x^{2}}{x^{2}\tan^{2}x}\notag\\ &= \lim_{x \to 0}\frac{\tan^{2}x - x^{2}}{x^{4}}\cdot\frac{x^{2}}{\tan^{2}x}\notag\\ &= \lim_{x \to 0}\frac{\tan^{2}x - x^{2}}{x^{4}}\cdot 1\notag\\ &= \lim_{x \to 0}\frac{\tan x - x}{x^{3}}\cdot \frac{\tan x + x}{x}\notag\\ &= \lim_{x \to 0}\frac{\tan x - x}{x^{3}}\cdot \lim_{x \to 0}\left(\frac{\tan x}{x} + 1\right)\notag\\ &= 2\lim_{x \to 0}\frac{\tan x - x}{x^{3}}\notag\\ &= 2\lim_{x \to 0}\frac{\sec^{2}x - 1}{3x^{2}}\text{ (via LHR)}\notag\\ &= \frac{2}{3}\lim_{x \to 0}\frac{\tan^{2}x}{x^{2}}\notag\\ &= \frac{2}{3}\notag \end{align}

Answer 4

I think this limit would be considerably easier using Taylor Series instead of LHR.

$ \frac{\sin^2(x)-x^2\cos^2(x)}{x^2\sin^2(x)} \approx \frac{(x-\frac{x^3}{6})^2-x^2(1-\frac{x^2}{2})^2}{x^2(x)^2}=\frac{\frac{2x^4}{3}+O(x^5)}{x^4} \rightarrow \frac{2}{3} $