Question #db815

Mar 5, 2017

${\lim}_{x \to 0} {\left(1 + 2 x\right)}^{\frac{2 x + 5}{x}} = {e}^{10}$

Explanation:

Write the function as:

${\left(1 + 2 x\right)}^{\frac{2 x + 5}{x}} = {\left({e}^{\ln} \left(1 + 2 x\right)\right)}^{\frac{2 x + 5}{x}} = {e}^{\frac{\ln \left(1 + 2 x\right) \left(2 x + 5\right)}{x}}$

Now consider the function:

$f \left(x\right) = \frac{\ln \left(1 + 2 x\right) \left(2 x + 5\right)}{x} = 2 \ln \left(1 + 2 x\right) + 5 \ln \frac{1 + 2 x}{x}$

we have that:

${\lim}_{x \to 0} 2 \ln \left(1 + 2 x\right) = 0$

while:

${\lim}_{x \to 0} \ln \frac{1 + 2 x}{x}$

is in the form $\frac{0}{0}$ so we can solve it using l'Hospital's rule:

${\lim}_{x \to 0} \ln \frac{1 + 2 x}{x} = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \ln \left(1 + 2 x\right)}{\frac{d}{\mathrm{dx}} x} = {\lim}_{x \to 0} \frac{2}{1 + 2 x} = 2$

Putting it togeher:

${\lim}_{x \to 0} 2 \ln \left(1 + 2 x\right) + 5 \ln \frac{1 + 2 x}{x} = 2 \cdot 0 + 5 \cdot 2 = 10$

Now, as ${e}^{x}$ is a continuous function we have:

${\lim}_{x \to 0} {e}^{f \left(x\right)} = {e}^{{\lim}_{x \to 0} f \left(x\right)}$

so that:

${\lim}_{x \to 0} {\left(1 + 2 x\right)}^{\frac{2 x + 5}{x}} = {\lim}_{x \to 0} {e}^{\frac{\ln \left(1 + 2 x\right) \left(2 x + 5\right)}{x}} = {e}^{10}$