Find $\lim_{x \to \infty} e^{2x \cdot \ln \frac{x+1}{x-2}}$ without l'Hospital.

Question

I have managed to get my limit to this state, but how should I procceed?

Thanks.

Answer 1

Hint: Suppose $$y=e^{2x\ln\frac{x+1}{x-2}}$$ then $$\ln y=2x\ln\frac{x+1}{x-2}=2x\ln(1+\frac{3}{x-2})=\ln(1+\frac{3}{x-2})^{2x}$$ or $$y=(1+\frac{3}{x-2})^{2x}$$ Now use the limit $$\lim_{n\to\infty}(1+\frac{1}{n})^n=e$$ to proceed.

Answer 2

$e^{2x\ln \frac {x+1}{x-2}}\\ (\frac {x+1}{x-2})^{2x}\\ (1 + \frac {3}{x-2})^{2x}$

$\lim_\limits{n\to\infty} (1+\frac 1n)^n = e\\ \lim_\limits{n\to\infty} (1+\frac xn)^n = e^x\\$

We have

$\lim_\limits{x\to\infty}(1 + \frac {3}{x-2})^{2x}$

If you want to be a little bit more thorough

$(1 + \frac {3}{x-2})^{2(x-2) + 4}$

The limit of the product is the product of the limits (if both limits are real numbers).

$\lim_\limits{x\to\infty}((1 + \frac {3}{x-2})^{x-2})^2\lim_\limits{x\to\infty}((1 + \frac {3}{x-2})^4)$

The first evaluates to $e^6$, the second evaluates to $1$

Answer 3

$\begin{align}\displaystyle\lim_{x \to \infty} e^{2x\ln \frac{x+1}{x-2}}&=\displaystyle\lim_{x \to \infty} e^{\ln\left(\frac{x+1}{x-2}\right)^{2x}}\\&=\displaystyle\lim_{x \to \infty}\left(\frac{x+1}{x-2}\right)^{2x}\\&=\displaystyle\lim_{x \to \infty}\left(\frac{(x-2)+3}{x-2}\right)^{2x}\\&=\displaystyle\lim_{x \to \infty}\left(1+\frac{3}{x-2}\right)^{2x}\\&=\displaystyle\lim_{x \to \infty}\left[\left(1+\frac{3}{x-2}\right)^{(x-2)+2}\right]^2\\&=\displaystyle\lim_{x \to \infty}\left[\left(1+\frac{3}{x-2}\right)^{(x-2)}\left(1+\frac{3}{x-2}\right)^2\right]^2\\&=\displaystyle\lim_{x \to \infty}\left[\left(1+\frac{3}{x-2}\right)^{(x-2)}\right]^2\left[\left(1+\frac{3}{x-2}\right)^2\right]^2\\&=\displaystyle\lim_{x \to \infty}\left[\left(1+\frac{3}{x-2}\right)^{(x-2)}\right]^2\displaystyle\lim_{x \to \infty}\left(1+\frac{3}{x-2}\right)^4\\&=\left[\displaystyle\lim_{x \to \infty}\left(1+\frac{3}{x-2}\right)^{(x-2)}\right]^2\displaystyle\lim_{x \to \infty}\left(1+\frac{3}{x-2}\right)^4\\&=\left[\displaystyle\lim_{(x-2) \to \infty}\left(1+\frac{3}{x-2}\right)^{(x-2)}\right]^2\displaystyle\lim_{x \to \infty}\left(1+\frac{3}{x-2}\right)^4\\&=(e^3)^2\cdot 1\\&=\boxed{\boxed{e^6}}\end{align}$