How to solve this limit without L'Hôpital's rule?

Question

$$\lim_{x\to\ + ∞} xe^{\frac{x-1}{x+1}}$$

Can you tell me how to solve this without L'Hôpital's rule? Subsituting $+∞$ I get $\frac{∞}{∞}$ in the exponent.

Answer 1

Observe that $$ xe^{\frac{x-1}{x+1}}>x,\quad x>1, $$ giving $$ \lim_{x\to\ + ∞} xe^{\frac{x-1}{x+1}}\ge \lim_{x\to\ + ∞}x. $$

Answer 2

$$ \ln(xe^{\frac{x-1}{x+1}})=\ln x+\frac{x-1}{x+1}\rightarrow\infty\quad\text{as} \quad x\rightarrow\infty $$

$$ \lim\limits_{x\rightarrow\infty}xe^{\frac{x-1}{x+1}}=e^{\lim\limits_{x\rightarrow\infty}\ln(xe^{\frac{x-1}{x+1}})}\rightarrow\infty $$

Answer 3

$$\lim_{x\to\ + ∞} {\frac{x-1}{x+1}}= 1$$ Thus $$\lim_{x\to\ + ∞} xe^{(x-1)/(x+1)}=\lim_{x\to\ + ∞} xe=+\infty$$

Or: $$\lim_{x\to\ + ∞} xe^{\frac{x-1}{x+1}}=\lim_{x\to\ + ∞} e^{\log(x\exp((x-1)/(x+1)))}=\lim_{x\to\ + ∞} e^{(x-1)/(x+1)+\log x}=e^{1+\infty}=+\infty$$