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

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?...

• It's not wrong to use the fact that $1^n \to 1$, but it's wholly unnecessary. The fact that $\frac{1}{e^{n^2x}} = \left(\frac 1 {e^x}\right)^{n^2}$ is just something algebraic. – T. Bongers Nov 27 '18 at 18:04

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

• I'm pretty fine with the fact that we apply $\frac{1}{e^{n^2 x}} = \left(\frac{1}{e^x}\right)^{n^2}$ as an algebraic transformation, but I still don't know if, and eventually when, is possible to substitute one expression with another due to the fact that they have the same limit, like I did substituting $1$ with $1^n$ – Alessio Martorana Nov 27 '18 at 22:09

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.

• The asker has literally said "calculating the easy limit … for $x < 0, x = 0$ and $x > 0$". Did you read the question and identify the parts of their work they are actually skeptical about? – T. Bongers Nov 27 '18 at 18:09
• @T.Bongers Yes you are right. I firstly didn't recognize the doubt properly. Now all should be fixed. Thanks – gimusi Nov 27 '18 at 18:17
• You still haven't made an effort to address the line that says "This is the passage I'm most uncertain of..." You're just repeating a computation that the asker has done without commenting on the validity of how they've rewritten something. Which was the real question. – T. Bongers Nov 27 '18 at 18:19
• @T.Bongers Ah ok, I add something of specific about that! – gimusi Nov 27 '18 at 18:21