Find $\lim_{x\to\infty} e^{\frac{3x}{x+1}}$

Question

I have a problem in solving this assignment question,pls help

Evaluate the limit of the sequence $(a_n)_{n=1 \to \infty}$ , if it exists:

$a_n = e^{\frac{3n}{n+1}}$

$\lim_{n\to\infty} e^{\frac{3n}{n+1}}$

Answer 1

The exponential is continous and clearly $\frac{3n}{n+1}$ goes to $3$ as $n\to\infty$.

Thus $\lim_n \exp{\frac{3n}{n+1}}=e^3$.

Answer 2

it is $$e^{\lim_{x \to \infty}\frac{3}{1+\frac{1}{x}}}=e^3$$

Answer 3

Theorem: Let $f$ be a continuous function at $c$. Then for every function $g$, $$\lim_{x\to c} f(g(x))=f\left(\lim_{x\to c}g(x)\right)$$ furthermore, if $f$ is continuous at $x$ for all $x>k$ for some $k$ then $$\lim_{x\to \infty} f(g(x))=f\left(\lim_{x\to\infty}g(x)\right)$$

Since $\lim_{x\to\infty}(3x)/(x+1)=3$, by this theorem the answer is $e^3$

Answer 4

$$\lim_{x\to\infty} e^{\frac{3x}{x+1}} =\lim_{x\to\infty} e^{3\frac{x}{x+1}} = e^{3\lim_{x\to\infty}\frac{x}{x+1}} = e^{3\cdot 1} = e^3$$