# L'Hospital's Rule Examples for Indeterminate Powers

Recall from the Indeterminate Forms page that:

If we have a limit in the form $\lim_{x\rightarrow a} f(x)^{g(x)}$, that is as $x \to a$, $f(x) \to 0$ and $g(x) \to 0$, then we have a limit of **Indeterminate Form of Type $0^0$**

If as $x \to a$, $f(x) \to \infty$ and $g(x) \to 0$, then we have a limit of **Indeterminate Form of Type $0^{\infty}$**.

If as $x \to a$, $f(x) \to 1$ and $g(x) \to \infty$, the we have a limit of **Indeterminate Form of Type $1^{\infty}$**.

## Example 1

**Evaluate the following limit using L'Hospital's rule: $\lim_{x \to 0^+} (4x + 1)^{\cot x}$.**

Let's first let $y = (4x + 1)^{\cot x}$. We can now take the natural logarithm of both sides and simplify this in order to obtain:

(1)Thus it follows that:

(2)We're really interested in only $\frac{\ln (4x + 1)}{\tan x}$. If we can evaluate $L = \lim_{x \to 0^+} \frac{\ln (4x + 1)}{\tan x}$, then $\lim_{x \to 0^+} (4x + 1)^{\cot x} = e^L$. Let's evaluate this limit:

(3)Notice that as $x \to 0^+$, $\ln(4x + 1) \to 0$, and $\tan x \to 0$. Hence we have an indeterminate form $\frac{0}{0}$. We can apply L'Hospital's rule:

(4)Hence it follows that since $\lim_{x \to 0^+} (4x + 1)^{\cot x} = e^L$, then:

(5)## Example 2

**Evaluate the following limit using L'Hospital's rule: $\lim_{x \to 0^+} (\cos x)^{\frac{1}{x^2}}$.**

Let $y = (\cos x)^{\frac{1}{x^2}}$. Thus $y = e^{\frac{\ln ( \cos x)}{x^2}}$.

(6)Let $L = \lim_{x \to 0^+} \frac{\ln ( \cos x)}{x^2}$ so that $\lim_{x \to 0^+} e^{\frac{\ln ( \cos x)}{x^2}} = e^L$. Let's now evaluate for $L$:

(7)As $x \to 0^+$, $\ln( \cos x) \to 0$ and $x^2 \to 0$, so we can apply L'Hospital's rule:

(8)Since as $x → 0^+$, $-\sin x \to 0$ and $2x\cos x \to 0$, then we can apply L'Hospital's rule again:

(9)Thus it follows that since $\lim_{x \to 0^+} e^{\frac{\ln ( \cos x)}{x^2}} = e^L$, then:

(10)