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Lets say I have: $$\lim _{x\to \:0^-}\left(e^{\frac{1}{x}}\right)$$

Is it allowed for me to do such a thing? $$\lim _{x\to 0^+}\left(e^{-\frac{1}{x}}\right)$$

or another example: $$\lim _{x\to -\infty }\left(e^{\frac{1}{x}}\right)$$ and change it to that limit: $$\lim _{x\to \infty }\left(e^{-\frac{1}{x}}\right)=\lim _{x\to \infty }\left(\frac{1}{e^{\frac{1}{x}}}\right)$$

Are the limit I wrote, is it allowed?

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  • $\begingroup$ The limits might be equal, but the functions are not. So it is not allowed. $\endgroup$ Commented Feb 22, 2022 at 15:48
  • $\begingroup$ What do you mean? why isnt it the same? As i am getting good values of limit when I do it. Do you have any example for me? it will help me understand it. $\endgroup$
    – user1028660
    Commented Feb 22, 2022 at 15:48
  • $\begingroup$ As I understand you say that $\lim\limits _{x\to \:0}\left(e^{\frac{1}{x}}\right)$ and $\lim\limits _{x\to \:0}\left(e^{-\frac{1}{x}}\right)$ do diverge. So both calculations are the same. But they are not. You consider two different functions. $\endgroup$ Commented Feb 22, 2022 at 15:53
  • $\begingroup$ Oh look what I wrote but, one at 0+ and other at 0- ( I wrote at (), I dont know how to type the 0+ and 0- in latex ) $\endgroup$
    – user1028660
    Commented Feb 22, 2022 at 15:56
  • $\begingroup$ If you are looking for $\lim\limits _{x\to \:0}\left(e^{\frac{1}{x}}\right)$, then you make a case decision:$\lim\limits _{x\to \:0^-}\left(e^{\frac{1}{x}}\right)$ and $\lim\limits _{x\to \:0^+}\left(e^{\frac{1}{x}}\right)$ $\endgroup$ Commented Feb 22, 2022 at 15:59

2 Answers 2

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Yes: changing the argument from $x\to -x$ is equivalent to reflecting the function wrt the $y$ axis, thus simply geometrically it's quite clear that $$\lim_{x\to x_0}f(x)=\lim_{x\to -x_0}f(-x)$$ with directions reversed if the limits are unilateral.

In general refer to this question Formal basis for variable substitution in limits

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  • $\begingroup$ Oh thanks!!!!!! $\endgroup$
    – user1028660
    Commented Feb 22, 2022 at 22:06
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the first point is right because : $\lim _{x\to \:0^-}f(x)=L$ means that : for all $\epsilon>0$ there exists $\delta>0$ such that : $$-\delta<x<0 \implies |f(x)-L|<\epsilon$$ which is equivalent to write that :
$$0<-x<\delta \implies |f(x)-L|<\epsilon$$ or $0<y<\delta \implies |f(-y)-L|<\epsilon$ (with $y=-x$)

which means that $\lim _{x\to \:0^+}f(-x)=L$

same thing for $\lim _{x\to \:0^-}f(x)=\infty \implies \lim _{x\to \:0^+}f(-x)=\infty$

and for $\lim _{x\to \infty}f(x)=L \implies \lim _{x\to -\infty}f(-x)=L$ for $L\in \overline{\mathbb{R}}$

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  • $\begingroup$ Thank you very much for the detailed answer :) $\endgroup$
    – user1028660
    Commented Feb 22, 2022 at 22:06

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