Limit question involving L'Hospitals rule

Question

I need help in solving the below question using L'Hospitals rule:

$$ \lim_{x \rightarrow 0} \left( \frac{1}{x^2}-\cot^2 x \right) $$

I'm getting infinity as the final answer. Thanks in advance.

Answer 1

$$ \lim_{x \rightarrow 0} \left( \frac{1}{x^2}-\cot^2 x \right) \\ = \lim_{x \rightarrow 0} \left( \frac{1}{x^2}-\frac{\cos^2 x}{\sin^2 x} \right) \\ = \lim_{x \rightarrow 0} \left( \frac{\sin^2 x}{x^2 \sin^2 x}-\frac{x^2 \cos^2 x}{x^2 \sin^2 x} \right) \\ = \lim_{x \rightarrow 0} \left( \frac{\sin^2 x - x^2 \cos x}{x^2 \sin^2 x} \right) \\ $$ Now you have a fraction where both numerator and denominator approach $0$ as $x \rightarrow 0$, so you can invoke L'Hospital's rule by taking the derivative of both top and bottom.

Answer 2

We use L'Hospital's Rule, although series are more natural.

Our function can be rewritten as $$\frac{1-x^2\cot^2 x}{x^2}.$$ Differentiate top and bottom. We get $$\frac{(-x^2)(2\cot x)(-\csc^2 x)-2x\cot^2 x}{2x},$$ which simplifies to $x\cot x\csc^2 x-\cot^2 x$.

Using the fact that $\cot x=\frac{\cos x}{\sin x}$, we find that we want $$\lim_{x\to 0}\cos x\left(\frac{x}{\sin^3 x}-\frac{\cos x}{\sin^2 x}\right).$$

The $\cos x$ part has limit $1$ as $x\to 0$. So we want $$\lim_{x\to 0} \frac{x-\sin x\cos x}{\sin^3 x}.$$

Use L'Hospital's Rule again. We want $$\lim_{x\to 0} \frac{1+\sin^2 x-\cos^2 x}{3\sin^2 x\cos x}.$$ We could continue to use L'Hospital's Rule. However, that would be a waste of time. For the top is simply $2\sin^2 x$, so the limit is $\frac{2}{3}$.