Find the limit as x goes to 0 of (x + e2x)1/x

Evaluate the limit.

$\lim_{x \to 0} \left( x + e^{2x} \right)^{\frac{1}{x}}.$

From the definition of the exponential we have

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

So, first we use the expansion of $e^x$ as $x \to 0$ (page 287 of Apostol) to write

$e^{2x} = 1 + 2x + o(x) \qquad \text{as} \quad x \to 0.$

Therefore, as $x \to 0$ we have

$\log (x + e^{2x}) = \log (x + 1 + 2x + o(x)) = \log (1 + 3x + o(x)).$

Now, since $3x + o(x) \to 0$ as $x \to 0$ we can use the expansion (again, page 287) of $\log (1+x)$ as $x \to 0$ to write

$\log (1 + 3x + o(x)) = 3x+o(x) + o(3x+o(x)) = 3x + o(x) \qquad \text{as} \quad x \to 0.$

Therefore, as $x \to 0$ we have

$\frac{1}{x} \log (x+e^{2x}) = \frac{1}{x}(3x + o(x)) = 3 + \frac{o(x)}{x}.$

So, getting back to the expression we started with,

$\lim_{x \to 0} \left( x + e^{2x} \right)^{\frac{1}{x}} = \lim_{x \to 0} e^{3 + \frac{o(x)}{x}} = e^3 \lim_{x \to 0} e^{\frac{o(x)}{x}}.$

But, as in the previous exercise (Section 7.11, Exercise #23) we know $\lim_{x \to 0} e^{\frac{o(x)}{x}} = 1$ . Hence,

$\lim_{x \to 0} \left( x + e^{2x} \right)^{\frac{1}{x}} = e^3.$

Stumbling Robot