Find the limit of (1 + 1/x)^(x^2 / e^x) as x approaches 0.

Step-by-step solution

1. 1

   Check where the expression is even defined. For small $x<0$ the base $1+\tfrac1x$ is negative (at $x=-0.1$ it equals $-9$), and a negative base raised to an arbitrary real power is undefined over the reals. So only the right-hand limit can be discussed, and any answer must be labelled as $x\to 0^{+}$.

2. 2

   Identify the indeterminate form. As $x\to 0^{+}$:

   $1+\frac1x\to+\infty,\qquad \frac{x^2}{e^x}\to\frac{0}{1}=0$

   This is $\infty^{0}$, which is genuinely indeterminate — it can come out as any positive number depending on how fast each part moves.

3. 3

   Take logarithms to convert the power into a product. Setting $L=\left(1+\tfrac1x\right)^{x^2/e^x}$,

   $\ln L=\frac{x^2}{e^x}\ln\left(1+\frac1x\right)$

   Logs are the standard tool for every $\infty^{0}$, $0^{0}$ or $1^{\infty}$ form: they turn the contest between base and exponent into a plain product.

4. 4

   Estimate the logarithm. For small positive $x$, $1+\tfrac1x\sim\tfrac1x$, so $\ln\left(1+\tfrac1x\right)\sim-\ln x$. Since $e^x\to 1$, the whole product behaves like

   $\ln L\sim x^2\ln\frac1x$

5. 5

   Use the standard limit $x^{a}\ln x\to 0$. Any positive power of $x$ beats the logarithm:

   $\lim_{x\to 0^{+}}x^2\ln\frac1x=0\ \Longrightarrow\ \lim_{x\to 0^{+}}\ln L=0$

6. 6

   Exponentiate back. Since $\ln L\to 0$ and $\exp$ is continuous,

   $\lim_{x\to 0^{+}}L=e^{0}=1$

   Numerically: at $x=0.01$ the value is $1.000457$ and at $x=10^{-5}$ it is $1.0000000012$ — converging to $1$ ✓. The two-sided limit does not exist, because the function is not defined to the left of $0$.