Evaluate L = lim as x to 0 (e^2x+x)^(1/x)

Evaluate $\displaystyle L = \lim_{x\to0}\bigl(e^{2x}+x\bigr)^{1/x}$.

Step 1: Take the natural logarithm

At $x = 0$ the base is $e^{0}+0 = 1$ while the exponent $\dfrac1x$ becomes infinite: the form is $1^{\infty}$. Write $L = e^{I}$ with

\[ I = \lim_{x\to0}\frac{\ln\bigl(e^{2x}+x\bigr)}{x}, \]

a quotient of the form $\dfrac00$, since $\ln1 = 0$.

Step 2: Apply l'Hôpital's rule

\[ \frac{d}{dx}\ln\bigl(e^{2x}+x\bigr) = \frac{2e^{2x}+1}{e^{2x}+x}, \qquad \frac{d}{dx}\bigl[x\bigr] = 1, \]

\[ I = \lim_{x\to0}\frac{2e^{2x}+1}{e^{2x}+x}. \tag{1} \]

The quotient in (1) is continuous at the origin, so substitute directly: \[I = \frac{2e^{0}+1}{e^{0}+0} = \frac{3}{1} = 3. \tag{2}\]

Step 3: Recover $L$

\[ L = e^{I} = e^{3}. \]

Step 4: Verification

\[e^{2x}+x = 1+3x+2x^{2}+O\!\left(x^{3}\right) \;\Longrightarrow\; \ln\bigl(e^{2x}+x\bigr) = 3x+O\!\left(x^{2}\right) \;\;\checkmark\]

\[\frac{\ln\bigl(e^{2x}+x\bigr)}{x} = 3+O(x) \longrightarrow 3 \;\;\checkmark\]

\[x = 0.001:\ 20.035472; \qquad x = 0.01:\ 19.597968; \qquad x = -0.01:\ 20.603090 \;\;\checkmark\]