Evaluate: $\lim_{n\to \infty}\left(\frac{a+n}{b+n}\right)^n$ I tried this way : $$\lim_{n\to \infty}\left(\frac{a+n}{b+n}\right)^n=\lim_{n\to \infty}\left(1+\frac{a-b}{b+n}\right)^n$$
We know $\lim_{n\to \infty}\left(1+\frac1u\right)^u=e$. therefor in this case we have: $$\lim_{n\to \infty}\left(1+\frac{a-b}{b+n}\right)^{\tfrac{b+n}{a-b}}=e$$
Here $n$ goes to infinity so we can ignore the number ($b$)added to it in numerator of the exponent. So $\lim_{n\to \infty}\left(1+\frac{a-b}{b+n}\right)^n=e^{a-b}$
Is my answer right? and is there any other approach to evaluate the limit?

Your answer is right but, following your idea, we should proceed as follows
$$\left(1+\frac{a-b}{b+n}\right)^n=\left[\left(1+\frac{a-b}{b+n}\right)^\tfrac{b+n}{a-b}\right]^{\tfrac{n(a-b)}{b+n}}$$
For a rigorous justification of this step you can refer to the following

*

*Why is $1^{\infty}$ considered to be an indeterminate form


*About $\lim \left(1+\frac {x}{n}\right)^n$ (useful also for the alternative approach)


*Proof of product rule for limits (relevant using that $f(x)^{g(x)}=e^{g(x)\log f(x)}$
