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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?...

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

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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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