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?...

  • $\begingroup$ 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. $\endgroup$ – 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 $$\lim_{n \to \infty} \frac{1}{e^{n^2 x}} = \lim_{n \to \infty} \left(\frac{1}{e^x}\right)^{n^2}$$

  • $\begingroup$ 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 $1^n$ $\endgroup$ – Alessio Martorana Nov 27 '18 at 22:09

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


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


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.

  • $\begingroup$ 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? $\endgroup$ – T. Bongers Nov 27 '18 at 18:09
  • $\begingroup$ @T.Bongers Yes you are right. I firstly didn't recognize the doubt properly. Now all should be fixed. Thanks $\endgroup$ – gimusi Nov 27 '18 at 18:17
  • $\begingroup$ 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. $\endgroup$ – T. Bongers Nov 27 '18 at 18:19
  • $\begingroup$ @T.Bongers Ah ok, I add something of specific about that! $\endgroup$ – gimusi Nov 27 '18 at 18:21

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