# What is the limit of (2^x -32)/(x-5 ) as x approaches 5?

Jun 24, 2016

$32 \ln \left(2\right) \approx 22.1807$

#### Explanation:

Use L'Hôpital's rule:

If functions $f$ and $g$ are differentiable on an open interval $I$ except possibly at $c \in I$ and all of the following hold:

• ${\lim}_{x \to c} f \left(x\right) = {\lim}_{x \to c} g \left(x\right) = 0$ or $\pm \infty$
• $g ' \left(x\right) \ne 0$ for all $x \in I \text{\} \left\{c\right\}$
• ${\lim}_{x \to c} \frac{f ' \left(x\right)}{g ' \left(x\right)}$ exists

Then:

${\lim}_{x \to c} f \frac{x}{g} \left(x\right) = {\lim}_{x \to c} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

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In our example, let $f \left(x\right) = {2}^{x} - 32$ and $g \left(x\right) = x - 5$.

Note that:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({2}^{x} - 32\right) = \frac{d}{\mathrm{dx}} \left({\left({e}^{\ln \left(2\right)}\right)}^{x} - 32\right)$

$= \frac{d}{\mathrm{dx}} \left({e}^{\ln \left(2\right) \cdot x} - 32\right) = \ln \left(2\right) {e}^{\ln \left(2\right) \cdot x} = \ln \left(2\right) \cdot {2}^{x}$

$g ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(x - 5\right) = 1$

These are both differentiable in the whole of $\mathbb{R}$, let alone any open interval we might choose containing $c = 5$.

• ${\lim}_{x \to 5} f \left(x\right) = {\lim}_{x \to 5} \left({2}^{x} - 32\right) = 32 - 32 = 0$
• ${\lim}_{x \to 5} g \left(x\right) = {\lim}_{x \to 5} \left(x - 5\right) = 5 - 5 = 0$
• $g ' \left(x\right) = 1 \ne 0$ for all $x \in \mathbb{R}$
• ${\lim}_{x \to 5} \frac{f ' \left(x\right)}{g ' \left(x\right)} = \frac{\ln \left(2\right) \cdot {2}^{5}}{1} = 32 \ln \left(2\right)$

Hence:

${\lim}_{x - 5} \frac{f \left(x\right)}{g \left(x\right)} = {\lim}_{x \to 5} \frac{f ' \left(x\right)}{g ' \left(x\right)} = 32 \ln \left(2\right) \approx 22.1807$

graph{((2^x-32)/(x-5)-y)((x-5)^2+(y-32ln(2))^2-0.1)=0 [-32.34, 32.6, -3.77, 28.7]}