Limit and L'Hopitals

Question

I'm having trouble with this problem.

$\lim{n \to \infty} (1+\frac 3n)^n$

My professor said to use a proof to figure out that the limit of the ln of the function is 3, but I can't figure out how to do it.

Answer 1

If you know that $\lim_{x \to \infty} (1+\frac{1}{x})^x = e$, then $$\lim_{n \to \infty}\left(1+\frac{3}{n}\right)^n=\lim_{n \to \infty} \left(1+\frac{1}{n/3}\right)^{n/3 \cdot 3} = e^3.$$

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If you don't know that, then try the following. \begin{align*} \lim_{n \to \infty} \ln \left(1+\frac{3}{n}\right)^n &= \lim_{n \to \infty} n \ln\left(1+\frac{3}{n}\right)\\ &= \lim_{n \to \infty} \frac{\ln\left(1+\frac{3}{n}\right)}{1/n} & \text{"$\infty/\infty$"}\\ &= \lim_{n \to \infty} \frac{\frac{-3/n^2}{1+\frac{3}{n}}}{-1/n^2} & \text{l'Hôpital}\\ &= \lim_{n \to \infty} \frac{3}{1+\frac{3}{n}}\\ &= 3. \end{align*}

Then, by continuity, $$\lim_{n \to \infty} \left(1+\frac{3}{n}\right)^n=\lim_{n \to \infty}e^{\ln \left(1+\frac{3}{n}\right)^n}=e^{\lim_{n to \infty} \ln \left(1+\frac{3}{n}\right)^n}=e^3.$$

Answer 2

$\left(1+\dfrac{3}{n}\right)^n = e^{n\ln \left(1+\frac{3}{n}\right)}$. We now find:

$\displaystyle \lim_{n \to \infty} n\ln\left(1+\dfrac{3}{n}\right) = \displaystyle \lim_{n\to \infty} 3\dfrac{\ln \left(1+\dfrac{3}{n}\right)}{\dfrac{3}{n}} = 3\displaystyle \lim_{x \to 0} \dfrac{\ln(1+x)}{x} = 3\displaystyle \lim_{x\to 0} \dfrac{1}{1+x} = 3 \Rightarrow \left(1+\dfrac{3}{n}\right)^n \to e^3$.