What is $e^{\lim\limits_{x \to \infty} 5x\ln (2 + \frac{3}{x})}$? I'm trying to find  $\lim\limits_{x \to \infty} \left (2 + \frac{3}{x}\right)^{5x}$ which should equal to $\infty$. But in the process of using $e^{\lim\limits_{x \to \infty} 5x\ln (2 + \frac{3}{x})}$, I got $\frac{15}{2}$ as an answer to the exponent.
This is what I did. 
Thank you.
 A: You make a mistake near the very beginning:
$$\lim_{x\to\infty}5x\ln(2+\frac3x)\ne\infty\cdot0$$
You mistakenly think that $\ln(2+\frac3x)\to\ln(1)=0$, but it actually approaches $\ln(2)$.  Thus, your L'Hospital's is invalid.  One should instead get
$$\lim_{x\to\infty}5x\ln(2+\frac3x)=\infty\cdot\ln(2)=\infty$$
A: $$\begin{align}\lim _{x\to \infty }\left(\:5x\ln\:\left(2\:+\:\frac{3}{x}\right)\right) &= \lim _{t\to 0}\left(\:\frac{5}{t}\:\cdot \:\ln\left(2\:+\:3t\right)\right) \\&= \lim _{t\to 0}\left(\:\frac{5}{t}\:\cdot \:\left(\ln \:\left(2\right)+\frac{3}{2}t+o\left(t\right)\right)\right) \\&= \lim _{t\to 0}\left(\:\frac{5\ln\left(2\right)}{t}\:+\frac{15}{2}\right) \\&= \infty\end{align}$$
EDIT
Is an approximation obtained by the Taylor formula to the first order....look this https://en.wikipedia.org/wiki/Taylor%27s_theorem
Taylor's formula:
$$\sum _{n=0}^{\infty }\:\frac{f^{\left(n\right)}\left(x_0\right)}{n!}\left(x-x_0\right)^n$$
So in our case$(x_0 = 0)$:
$$\ln(2+3x)=\ln \left(2\right)+\frac{\frac{3}{2}}{1!}x+\frac{-\frac{9}{4}}{2!}x^2+\frac{\frac{27}{4}}{3!}x^3+\frac{-\frac{243}{8}}{4!}x^4+\ldots $$
but it is enough to stop at the first order.
