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(1+1/n)^(n^2)

Limit of the function (1+1/n)^(n^2)

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            / 2\
            \n /
     /    1\    
 lim |1 + -|    
n->oo\    n/    
$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n^{2}}$$
Limit((1 + 1/n)^(n^2), n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n^{2}}$$
transform
do replacement
$$u = \frac{n}{1}$$
then
$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n^{2}}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u^{2}}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u^{2}}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{u}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{u} = e^{u}$$

The final answer:
$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n^{2}} = \infty$$
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
The graph
Rapid solution [src]
oo
$$\infty$$
Other limits n→0, -oo, +oo, 1
$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n^{2}} = \infty$$
$$\lim_{n \to 0^-} \left(1 + \frac{1}{n}\right)^{n^{2}} = 1$$
More at n→0 from the left
$$\lim_{n \to 0^+} \left(1 + \frac{1}{n}\right)^{n^{2}} = 1$$
More at n→0 from the right
$$\lim_{n \to 1^-} \left(1 + \frac{1}{n}\right)^{n^{2}} = 2$$
More at n→1 from the left
$$\lim_{n \to 1^+} \left(1 + \frac{1}{n}\right)^{n^{2}} = 2$$
More at n→1 from the right
$$\lim_{n \to -\infty} \left(1 + \frac{1}{n}\right)^{n^{2}} = 0$$
More at n→-oo
The graph
Limit of the function (1+1/n)^(n^2)