# Find the limit using l'Hopital's Rule

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.

Find $\lim_{x\to \infty}\left(1+\frac{a}{x}\right)^{bx}$

I know to take the natural log of both sides, which would give you $bx \cdot \ln\left(1+\frac{a}{x}\right)$ but I'm not sure where to go from there. Can anyone help?

Write $$y=\lim_{x \rightarrow \infty}\frac{In(1+a/x)}{\frac{1}{bx}}$$ and keep applying L' Hopital's rules.

You will eventually reach $$y=\lim_{x \rightarrow \infty}\frac{a}{b(1+\frac{a}{x})}$$

$$L=\lim_{x\rightarrow \infty }(1+\frac{a}{x})^{bx}=\lim_{x\rightarrow \infty }((1+\frac{a}{x})^{x})^b=....$$

Substitute $y = \frac xa$. And it's easy to notice that $y \to \infty$, as $x \to \infty$. So we have that:

$$\lim_{x \to \infty} \left(1 + \frac ax\right)^{bx} = \lim_{y \to \infty} \left(1 + \frac 1y\right)^{aby} = \left(\lim_{y \to \infty} \left(1 + \frac 1y\right)^{y}\right)^{ab} = e^{ab}$$

where we used the continuity of the exponential function and the well-known fact that $\lim_{x \to \infty} \left(1 + \frac 1x \right)^{x} = e$

You can do a substitution, $t=1/x$, so you get (after taking the logarithm), $$\lim_{t\to0^+}b\frac{\ln(1+at)}{t}= b\lim_{t\to0^+}\frac{\ln(1+at)-0}{t-0}$$ Does this ring a bell?

$\lim_{x\to \infty}\left(1+\frac{a}{x}\right)^{bx}$

Just to be different

binomial expansion

$\lim_{x\to \infty}\left(1+bx\frac{a}{x} + \frac {bx(bx-1)}{2!} (\frac ax)^2 + \frac {bx(bx-1)(bx-2)}{3!} (\frac ax)^3 \cdots\right)\\ \sum_\limits{n=0}^\infty \frac {(ab)^n}{n!} = e^{ab}$

• Doug, how do you propose justifying the interchange of the limit and the series? Neither the Weierstrass M-Test nor the DCT is likely a part of the OP's repertoire. Oct 10, 2016 at 23:10