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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?

b00n heT
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  • The limits might be equal, but the functions are not. So it is not allowed. – callculus42 Feb 22 '22 at 15:48
  • 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. –  Feb 22 '22 at 15:48
  • 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. – callculus42 Feb 22 '22 at 15:53
  • 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 ) –  Feb 22 '22 at 15:56
  • 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 $\lim\limits _{x\to :0^+}\left(e^{\frac{1}{x}}\right)$ – callculus42 Feb 22 '22 at 15:59
  • No no, you didnt understand what I mean... look at the question again, my question, look regarding the limit of infinity, you will understand better, dont mind the limit of 0, it will just confuse you. The third limit I wrote, is it equal to the last two? look what I wrote and did. –  Feb 22 '22 at 16:01
  • Look only the last 3 limit of the X goes to infinity. dont mind the x goes to 0. –  Feb 22 '22 at 16:02
  • I would change the variable: $\lim\limits_{x\to -\infty }\left(e^{\frac{1}{x}}\right)$ Then $x=-y:$ $ \ \lim\limits_{y\to \infty }\left(e^{-\frac{1}{y}}\right)$ – callculus42 Feb 22 '22 at 16:14
  • ohh, that is a great idea... that way I can do it easier. The only reason I want to do this is because, lets say I have that limit you wrote, but when X goes to 0 minus, its confusing and I have made lots of mistakes because its zero + minus, and its pretty hard seeing it.. thanks –  Feb 22 '22 at 17:23

2 Answers2

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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

b00n heT
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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 $L\in \overline{\mathbb{R}}$

Olivier
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