Find limit of function.

Question

The problem: Find $\displaystyle \lim_{x\to 2}\left(\frac{2^x-4}{x-2}\right)$.

I have no idea how to start.

$t=2^x$ and I receive:

$\displaystyle \lim_{t\to 4}\left( \frac{t-4}{\log_2 t -2} \right)$ but it doesn't help.

Answer 1

One way

$$\lim _{ x\to 2 } \frac { 2^{ x }-4 }{ x-2 } \overset { L'Hospital }{ = } \lim _{ x\to 2 } \frac { 2^{ x }\ln { 2 } }{ 1 } =4\ln { 2 } $$

Second way since $\lim _{ x\rightarrow 0 }{ \frac { { a }^{ x }-1 }{ x } =\ln { a } } $ we have (by substitution $x-2=t$)

$$\\ \lim _{ x\to 2 } \frac { 2^{ x }-4 }{ x-2 } =\lim _{ t\rightarrow 0 } \frac { 2^{ t+2 }-4 }{ t } =\lim _{ t\rightarrow 0 } 4\frac { 2^{ t }-1 }{ t } =4\ln { 2 } $$

Answer 2

Notice that $2^x$ is differentiable everywhere, so $$ \lim_{x \to 2} \frac{2^x - 4}{x - 2} = \left.\frac{\mathrm d}{\mathrm d x} 2^x \right|_2 = 4 \log 2 $$

Answer 3

Notice that

$$\frac d{dx}2^x=\lim_{h\to x}\frac{2^h-2^x}{h-x}$$

which is one of the definitions of the derivative. The derivative of $2^x$ at $x=2$ is given by

$$\left.\frac d{dx}2^x\right|_{x=2}=\lim_{h\to2}\frac{2^h-4}{h-2}$$

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Another way to show this is by some brief rearranging:

$$\begin{align}\lim_{h\to2}\frac{2^h-4}{h-2}&=4\lim_{h\to2}\frac{2^{h-2}-1}{h-2}\\&=4\lim_{h\to1}\frac{2^{h-1}-1}{h-1}\\&=4\int_1^2\frac1xdx\\&=4\ln(2)\end{align}$$

Since $\displaystyle\int_1^r\frac1{x^n}dx=\frac{r^{n-1}}{n-1}-\frac{1}{n-1}=\frac{r^{n-1}-1}{n-1}$ by power rule, where we have $r=2$ and $n\to1$.