# Limit - Minus and Plus changing ( forbidden \ allowed? )

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?

• The limits might be equal, but the functions are not. So it is not allowed. Commented Feb 22, 2022 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.
– user1028660
Commented Feb 22, 2022 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. Commented Feb 22, 2022 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 )
– user1028660
Commented Feb 22, 2022 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 $\lim\limits _{x\to \:0^+}\left(e^{\frac{1}{x}}\right)$ Commented Feb 22, 2022 at 15:59

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

• Oh thanks!!!!!!
– user1028660
Commented Feb 22, 2022 at 22:06

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 : $$-\delta which is equivalent to write that :
$$0<-x<\delta \implies |f(x)-L|<\epsilon$$ or $$0 (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 $$L\in \overline{\mathbb{R}}$$

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