calculate limit $\lim_{n\rightarrow\infty} \frac{1}{e^{n^2x}}$, x varying in real numbers

Question

Consider the calculation of the following limit: $$\lim_{n\rightarrow\infty}\frac{1}{e^{n^2x}}$$

I did these passages... Could you tell me if the whole resolution is formally right?

$$\lim_{n\rightarrow\infty}\frac{1}{e^{n^2x}} = \frac{\lim_{n\rightarrow\infty} 1}{\lim_{n\rightarrow\infty} e^{n^2x}}$$

(This is the passage I'm most uncertain of...)

I try to apply the famous limit: $\lim_{n\rightarrow\infty} 1^n = 1$, so I substitute 1 with $1^n$

$$= \frac{\lim_{n\rightarrow\infty} 1^n}{\lim_{n\rightarrow\infty} e^{n^2x}}$$

I do substitute 1 with $1^n$ again

$$\frac{\lim_{n\rightarrow\infty} (1^n)^n}{\lim_{n\rightarrow\infty} e^{n^2x}} = \lim_{n\rightarrow\infty} \frac{1^{n^2}}{e^{n^2 x}} = \lim_{n\rightarrow\infty} (\frac{1}{e^x})^{n^2}$$

At this point I substitute $n^2 = n'$ and study the base of the power ($\frac{1}{e^x}$) at the varying of x, calculating the easy limit $\lim_{n'\rightarrow\infty} (\frac{1}{e^x})^{n'}$for $x<0, x=0$ and $x>0$

It's all correct or there are some errors/imprecisions in the calculation? I'm very uncertain on the fact that I can, formally, substitute $1^n$ to $1$ only because of $\lim_{n\rightarrow\infty} 1^n = 1$... Is it possible to substitute one expression with another only because they have the same limit?...

Answer 1

You're making things more complicated than necessary.

For any $n$, we have $\frac{1}{e^{n^2 x}} = \left(\frac{1}{e^x}\right)^{n^2}$. Taking the limit of both sides yields $$\lim_{n \to \infty} \frac{1}{e^{n^2 x}} = \lim_{n \to \infty} \left(\frac{1}{e^x}\right)^{n^2}$$

Answer 2

Regarding your main doubt note that $\forall n$ and $\forall x$ the following identity holds

$$\frac{1}{e^{n^2x}}=\left(\frac{1}{e^{x}}\right)^{n^2}$$

and we don't need to use limit concept for that, it is indeed an algebraic identity, and now passing to the limit we obtain

$$\lim_{n\rightarrow\infty}\frac{1}{e^{n^2x}}=\lim_{n\rightarrow\infty}\left(\frac{1}{e^{x}}\right)^{n^2}$$

Then we can simply distinguish the cases

• $x>0 \implies \frac{1}{e^{n^2x}}=\left(\frac{1}{e^{x}}\right)^{n^2}\to 0\quad$ since $\frac{1}{e^{x}}<1$

• $x=0 \implies \frac{1}{e^{n^2x}}=1$

• $x<0\implies \frac{1}{e^{n^2x}}=\left(\frac{1}{e^{x}}\right)^{n^2}\to \infty\quad$ since $\frac{1}{e^{x}}>1$

and evaluate the limit for each one.