# lim_(xrarr0)(e^(2x)+x)^(1//x)= ?

Nov 25, 2016

${\lim}_{x \rightarrow 0} {\left({e}^{2 x} + x\right)}^{1 / x} = {e}^{3}$

#### Explanation:

$L = {\lim}_{x \rightarrow 0} {\left({e}^{2 x} + x\right)}^{1 / x}$

To undo powers like this in limits, take the natural logarithm of both sides.

$\ln \left(L\right) = \ln \left({\lim}_{x \rightarrow 0} {\left({e}^{2 x} + x\right)}^{1 / x}\right)$

The logarithm, being a continuous function, can be moved inside the limit.

$\ln \left(L\right) = {\lim}_{x \rightarrow 0} \ln \left({\left({e}^{2 x} + x\right)}^{1 / x}\right)$

Move the power outside of the logarithm using log rules.

$\ln \left(L\right) = {\lim}_{x \rightarrow 0} \ln \frac{{e}^{2 x} + x}{x}$

Notice that as $x \rightarrow 0$, we have this limit in the indeterminate form $\frac{0}{0}$. That means L'Hospital's rule applies, and we can take the derivatives of the numerator and denominator separately in the limit.

$\ln \left(L\right) = {\lim}_{x \rightarrow 0} \frac{\frac{d}{\mathrm{dx}} \left(\ln \left({e}^{2 x} + x\right)\right)}{\frac{d}{\mathrm{dx}} \left(x\right)}$

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

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

We can now evaluate the limit.

$\ln \left(L\right) = \frac{2 {e}^{0} + 1}{{e}^{0} + 0}$

$\ln \left(L\right) = \frac{2 + 1}{1}$

$\ln \left(L\right) = 3$

$L = {e}^{3}$

Nov 25, 2016

${e}^{3}$
${\left({e}^{2 x} + x\right)}^{\frac{1}{x}} = {e}^{2} {\left(1 + \frac{x}{e} ^ \left(2 x\right)\right)}^{\frac{1}{x}}$ now, making $y = \frac{x}{e} ^ \left(2 x\right)$
${\left({e}^{2 x} + x\right)}^{\frac{1}{x}} = {e}^{2} {\left(1 + y\right)}^{\frac{1}{y {e}^{2 x}}} = {e}^{2} {\left({\left(1 + y\right)}^{\frac{1}{y}}\right)}^{\frac{1}{{e}^{2 x}}}$
Now ${\lim}_{x \to 0} \implies {\lim}_{y \left(x\right) \to 0}$ so
${\lim}_{x \to 0} {\left({e}^{2 x} + x\right)}^{\frac{1}{x}} = {e}^{2} {\lim}_{x \to 0} {\left({\lim}_{y \to 0} {\left(1 + y\right)}^{\frac{1}{y}}\right)}^{\frac{1}{e} ^ \left(2 x\right)} = {e}^{2} \times e = {e}^{3}$