1. Check if the limit is in indeterminate form. 2. Verify conditions: $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$ or both approach $\pm \infty$. 3. Apply L'Hôpital's Rule: $\lim_{x\to a}\frac{f'(x)}{g'(x)}$. 4. Evaluate the new limit.

1. Check indeterminate form: $\frac{0}{0}$. 2. Apply L'Hopital's Rule: $\lim_{x \to 0} \frac{3\cos(3x)}{1}$. 3. Evaluate: $3\cos(0) = 3$.

1. Check indeterminate form: $\frac{\infty}{\infty}$. 2. Apply L'Hôpital's Rule: $\lim_{x\to \infty} \frac{6x}{14x}$. 3. Simplify: $\frac{6}{14} = \frac{3}{7}$.

An expression whose limit cannot be evaluated directly, such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

Limits of indeterminate forms.

First, verify that the limit is in an indeterminate form, either $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

Show that the limit of the numerator and the limit of the denominator both approach 0 or both approach $\pm \infty$.

It means showing that both $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$ or that both limits approach $\pm \infty$.

If after applying L'Hôpital's Rule once, the limit is still in an indeterminate form, you can apply it again.