L'Hospital's Rule and Indeterminate Forms — Question 2

Question 2

Evaluate the limit: $$\lim_{x \to 0^+} \left( \frac{1}{x} - \frac{1}{\sin x} \right)$$

Question 2 - Solution

We observe that both $\frac{1}{x} \to \infty$ and $\frac{1}{\sin x} \to \infty$ as $x \to 0^+$, so the expression is of the indeterminate form: $$\infty - \infty$$

We first combine into a single fraction: $$\frac{1}{x} - \frac{1}{\sin x} = \frac{\sin x - x}{x \sin x}$$

Now the limit becomes: $$\lim_{x \to 0^+} \frac{\sin x - x}{x \sin x}$$

As $x \to 0^+$, both numerator and denominator → 0. This is now a $\frac{0}{0}$ indeterminate form, so we apply L’Hospital’s Rule:

$$\lim_{x \to 0^+} \frac{\sin x - x}{x \sin x} = \lim_{x \to 0^+} \frac{\cos x - 1}{\sin x + x \cos x}$$

Now evaluate the new limit:

- Numerator: $\cos x - 1 \to 0$ - Denominator: $\sin x + x \cos x \to 0$

Still indeterminate! Apply L’Hospital’s Rule again:

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

Now evaluate: - Numerator: $-\sin x \to 0$ - Denominator: $2 \cos x - x \sin x \to 2$

So the final result is: $$\boxed{0}$$