# Evaluate $\lim_{s\to0}\frac{s\sqrt{e^s}}{e^s-1}$ without using L'Hospital rule.

Evaluate $\lim_{s\to0}\frac{s\sqrt{e^s}}{e^s-1}$ without using L'Hospital rule.

I've done this: Since $\sqrt{e^s}\to1$as $s\to1$, so $\lim_{s\to0}\frac{s\sqrt{e^s}}{e^s-1}=\lim_{s\to0}\frac{s}{e^s-1}=1$

Are my steps correct? Is step 1 to step 2 available? Thank you.

• What's your justification for the last limit being 1 without using L'Hopital? – Kevin Arlin Aug 21 '12 at 7:49
• @KevinCarlson: Usually, it follows straightforward from $(1+x)^\frac1x \to \mathrm e$ with $x\to 0$. To Jason: yes, since limits of both terms do exist, the limit of the product exists and equals to the product of limits. – Ilya Aug 21 '12 at 7:54
• @Ilya $(1+x)^{1/x} \to e$ as $x \to 0$! – Mercy King Aug 21 '12 at 7:57
• @Mercy: sorry, that was a typo - I was already thinking of equivalent pairs like $\log(1+x)\sim x$. – Ilya Aug 21 '12 at 7:58
• Hint for an alternative derivation (yours is right as shown by Ilya) : $\displaystyle \frac{s\sqrt{e^s}}{e^s-1}=\frac{s}{e^{\frac s2}-e^{-\frac s2}}$ and since the $\sinh\cdots$ – Raymond Manzoni Aug 21 '12 at 8:26

Well, indeed you are right: you can prove it just based on the limit $\lim\limits_{x\to 0}(1+x)^{1/x} = \mathrm e$. By taking the logarithm of both parts, you obtain $\lim\limits_{x\to 0}\frac{\log(1+x)}{x} = 1$. If you put $s =\log(1+x)$ then you get $$\lim\limits_{s\to 0}\frac{s}{\mathrm e^s-1} = 1$$ as you mentioned. Now, if you have $\lim\limits_{s\to 0}f(s) = F$ and $\lim\limits_{s\to 0}g(s) = G$ then $$\lim\limits_{s\to 0}f(s)g(s) = FG$$ and the limit in LHS exists - that's why you can go from step 1 to step 2 in your proof.

• @jasoncube: you are welcome – Ilya Aug 21 '12 at 14:07
• One could even make again the substitution $s= log(t+1)$ and derives again at the desired expression $\frac{\log(1+t)}{t}$ – Imago Oct 10 '17 at 13:23