Limit of $\lim_{x \rightarrow 0^+} e^{\frac{x}{x}}$

by Ski Mask   Last Updated January 17, 2018 16:20 PM

If I need to find $$\lim_{x \rightarrow 0^+} e^{\frac{x}{x}}$$ I know that since $e$ is a constant I can take it out and make the equation $$e^{\lim_{x \rightarrow 0^+} {\frac{x}{x}}}$$ Then use L'Hopital's Rule and differentiate $\frac{(x)'}{(x)'}=1$ and get that $e^{\lim_{x\rightarrow 0^+}1}=e^1$. Is my approach correct?

Answers 2

Note simply because: $$\frac{x} {x} =1$$ we have: $$\lim_{x \to 0^+} e^{\frac{x} {x}} = \lim_{x \to 0^+} e^1 = e$$

January 17, 2018 16:03 PM

note that $$\frac{x}{x}=1$$ if $x\neq 0$ so the searched limit is $e^{1}=e$

Dr. Sonnhard Graubner
Dr. Sonnhard Graubner
January 17, 2018 16:04 PM

Related Questions

Evaluate a limit using l'Hospital rule

Updated August 31, 2017 10:20 AM