Evaluate $\lim_{n\to \infty} \left(1+\frac{a}{n^2}\right)^\frac{n}{b}$ I stuck on calculation of $\lim_{n\to \infty }\left(1+\frac{a}{n^2}\right)^{\frac{n}{b}}$ where $a,b$ are constants. What is the trick to do it?
 A: HINT
$$\left(1+\frac{a}{n^2}\right)^\frac{n}{b}=e^{\frac n b \log \left(1+\frac{a}{n^2}\right) }$$
and
$$\frac n b \log \left(1+\frac{a}{n^2}\right)\sim\frac a b\frac1n $$
or as an alternative
$$\left(1+\frac{a}{n^2}\right)^\frac{n}{b}=\left[\left(1+\frac{a}{n^2}\right)^\frac{n^2}{a}\right]^{\frac{a}{b}\frac1n}$$
A: If the following limit exists, it is equal to the requested one:
$$
\lim_{x\to0^+}(1+ax^2)^{\tfrac{1}{bx}}
$$
By the continuity of exponential and logarithm, you can compute
$$
\lim_{x\to0^+}\log\Bigl(1+ax^2)^{\tfrac{1}{bx}}\Bigr)=
\lim_{x\to0^+}\frac{\log(1+ax^2)}{bx}=
\lim_{x\to0^+}\frac{ax^2+o(x^2)}{bx}=0
$$
So your given limit is $e^0=1$.
A: $$\begin{align} \lim_{n\to\infty} e^{\log{\left(\dfrac{a}{n^2}+1\right)}} \\ \\= \lim_{n\to\infty} e^\dfrac{{n\log{\left(\dfrac{a}{n^2}+1\right)}}}{b} \\ \\ = \lim_{e^n\to\infty} n\log{\left(\dfrac{a}{n^2}+1\right)}{b} \\ \\ = e^{\dfrac{\lim_{n\to\infty} n\log\left(\dfrac{a}{n^2}+1\right)}{\lim_{n\to\infty}  b}}\end{align}$$
Then apply l'Hopital's rule:
$$\begin{align} e^{\lim_{n\to\infty} \dfrac{2an}{a+n^2}} b \\ \\= e^{\dfrac{\lim_{n\to\infty} \dfrac{n}{a+n^2}2a}{b}} \\ \\ = e^{\dfrac{2 \cdot 0 \cdot a}{b}} \\ \\ e^{\frac{0}{b}} \implies \lim_{n\to\infty} e^{\log{\left(\dfrac{a}{n^2}+1\right)}} = 1\end{align}$$
A: We prove a general result and then specialize.
Lemma If $c_j\to 0$, $a_j\to \infty$ and $c_ ja_j\to \lambda$, then $(1+c_j)^{a_j}\to e^{\lambda}$.
Proof Take the logarithm and note that $$\lim_{j\to \infty}a_j\log(1+c_j)
=\lim_{j\to \infty}a_jc_j\times\lim_{j\to \infty}\frac{\log(1+c_j)-\log(1)}{c_j-0}=\lambda\times 1=\lambda$$
where we used the definition of the derivative on the second limit.$\blacksquare$
In our case $c_j=a/j^2$ and $a_j=j/b$, so $c_ja_j\to 0$ whence
$$
\lim_{j\to\infty}\left(1+\frac{a}{j^2}\right)^{j/b}=e^0=1
$$
A: Since $f(x) =x^{k}, k\in\mathbb{R} $ is continuous on $(0,\infty)$ we have the following result:

Lemma: If $x_n\to L>0$ then $x_n^{k} \to L^{k} $ as $n\to\infty$.

And next is the famous (on MSE) result courtesy Thomas Andrews:

Theorem: If $n(a_n-1)\to 0$ then $a_n^n\to 1$ as $n\to\infty$.

Let $a_n=1+(a/n^2)$ so that $n(a_n-1)\to 0$ and hence $a_n^n=(1+(a/n^2))^n\to 1$ and by the lemma stated in the beginning (with $k=1/b$) we have $(1+(a/n^2))^{n/b}\to 1$.
No properties of logarithm or exponential functions are needed. 
A: Hint:

If $\lim_{x\to a} f(x)=1$ and $\lim_{x\to a} g(x)\to \infty$ then $$\lim_{x\to a} f(x)^{g(x)}=e^{\lim_{x\to a} g(x)[f(x)-1]}$$

The above statement can be proved by writing $f(x)^{g(x)}=e^{g(x)\ln (f(x))}$ and knowing that for a continuous function $h(x)$ at given point $\lim_{x\to a} h(I(x))=h(\lim_{x\to a} I(x))$ and it is quite well obvious that $e^x$ is continuous  over all real numbers and hence limit can enter the function easily. 
In your case $f(x)=\left(1+\frac {a}{n^2}\right) $ and $g(x)=\frac nb$
