# Question #7bf46

Feb 5, 2017

${\lim}_{x \to 0} {\left({e}^{x} + 30 x\right)}^{\frac{1}{x}} = {e}^{31}$

#### Explanation:

Consider the function:

$f \left(x\right) = {\left({e}^{x} + 30 x\right)}^{\frac{1}{x}}$

and take its logarithm:

$\ln f \left(x\right) = \ln \left({\left({e}^{x} + 30 x\right)}^{\frac{1}{x}}\right)$

using the properties of logarithm this is:

$\ln f \left(x\right) = \frac{1}{x} \ln \left({e}^{x} + 30 x\right)$

Now evaluate:

${\lim}_{x \to 0} \ln f \left(x\right) = {\lim}_{x \to 0} \ln \frac{{e}^{x} + 30 x}{x}$

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

${\lim}_{x \to 0} \ln f \left(x\right) = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \ln \left({e}^{x} + 30 x\right)}{\frac{d}{\mathrm{dx}} x} = {\lim}_{x \to 0} \frac{{e}^{x} + 30}{{e}^{x} + 30 x} = 31$

Since $\ln x$ is a continuous function in $\left(0 , + \infty\right)$:

$31 = {\lim}_{x \to 0} \ln f \left(x\right) = \ln \left({\lim}_{x \to 0} f \left(x\right)\right)$

so that:

${\lim}_{x \to 0} f \left(x\right) = {e}^{31}$