The trick for powers A to the B

When a limit has the form $\bigl[A(x)\bigr]^{B(x)}$ with an indeterminate form such as $1^\infty$, $\infty^0$ or $0^0$, the trick is to rewrite the power as an exponential:

$A(x)^{B(x)} = e^{\,B(x)\cdot\ln A(x)}.$

The limit of the exponent $B(x)\cdot\ln A(x)$ is usually an “ordinary” indeterminate form ($0\cdot\infty$ or similar) that is solved with the methods seen. Then one applies the continuity of the exponential.

Example — The limit that gives $e$

$\lim_{x\to 0^+}(1+x)^{1/x} = \lim e^{\frac{1}{x}\cdot\ln(1+x)}.$ We use $\ln(1+\square)\sim\square$ with $\square = x\to 0$: $\frac{1}{x}\cdot\ln(1+x) \sim \frac{1}{x}\cdot x = 1.$ Hence the limit is $e^1 = \boxed{e}$.

Example — Base tending to 1, exponent to infinity

$\lim_{x\to+\infty}\left(\frac{x+1}{x-1}\right)^x = e^{\,x\cdot\ln\frac{x+1}{x-1}} = e^{\,x\cdot\ln\left(1+\frac{2}{x-1}\right)}.$ With $\square = \dfrac{2}{x-1}\to 0$ we have $\ln(1+\square)\sim\square$, hence $x\cdot\frac{2}{x-1}\to 2.$ The limit is $\boxed{e^2}$.

Links

Topics: Limits
Concepts: Indeterminate forms · Equivalent functions
Skills: Calculating limits · Using formulae