L'Hospital's Rule Examples for Indeterminate Powers

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)
\begin{align} \ln y = \ln ((4x + 1)^{\cot x}) \\ \ln y = \cot x \cdot \ln (4x + 1) \\ \ln y = \frac{\ln (4x + 1)}{\tan x} \\ y = e^{\frac{\ln (4x + 1)}{\tan x}} \end{align}

Thus it follows that:

(2)
\begin{align} \lim_{x \to 0^+} (4x + 1)^{\cot x} = \lim_{x \to 0^+} e^{\frac{\ln (4x + 1)}{\tan x}} \end{align}

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)
\begin{align} L = \lim_{x \to 0^+} \frac{\ln (4x + 1)}{\tan x} \end{align}

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)
\begin{align} L = \lim_{x \to 0^+} \frac{\ln (4x + 1)}{\tan x} = \lim_{x \to 0^+} \frac{\frac{4}{4x + 1}}{\sec ^2 x} \\ L = \lim_{x \to 0^+} \frac{4 \cos ^2 x}{4x + 1} \\ L = 4 \end{align}

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

(5)
\begin{align} \lim_{x \to 0^+} (4x + 1)^{\cot x} = e^4 \end{align}

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)
\begin{align} \lim_{x \to 0^+} (\cos x)^{\frac{1}{x^2}} = \lim_{x \to 0^+} e^{\frac{\ln ( \cos x)}{x^2}} \end{align}

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)
\begin{align} L = \lim_{x \to 0^+} \frac{\ln ( \cos x)}{x^2} \end{align}

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

(8)
\begin{align} L = \lim_{x \to 0^+} \frac{-\sin x}{2x \cos x} \end{align}

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

(9)
\begin{align} L = \lim_{x \to 0^+} \frac{-\cos x}{2\cos x - 2x \sin x} \\ L = -\frac{1}{2} \end{align}

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

(10)
\begin{align} \lim_{x \to 0^+} (\cos x)^{\frac{1}{x^2}} = e^{\frac{-1}{2}} = \frac{1}{\sqrt{e}} \end{align}