# Why I don't get the right result for the limit $\lim _{x\to \infty }\frac{\left(\left(1-x\right)e^{\frac{1}{2x+2}}\right)}{x}$?

Good evening to everyone, I have the following limit $\lim _{x\to \infty }\frac{\left(\left(1-x\right)e^{\frac{1}{2x+2}}\right)}{x}$ the result of it is $-1$ but my calculations give another result $$\lim _{x\to \infty }\frac{\left(\left(1-x\right)e^{\frac{1}{2x+2}}\right)}{x} =^H \lim _{x\to \infty }\frac{\frac{d}{dx}\left(\left(1-x\right)e^{\frac{1}{2x+2}}\right)}{\frac{d}{dx}x} = \lim _{x\to \infty }\left(e^{\frac{1}{2x+2}}+\left(1-x\right)2e^{\frac{1}{2x+2}}\right) = -\infty$$

• The derivative of $\exp({1/(2x+2)})$ is not $2\exp({1/(2x+2)})$ Jun 15 '16 at 20:36

How did you differentiate that? I get:

$$\left[(1-x)e^{\frac1{2x+2}}\right]'=-e^{\frac1{2x+2}}-\frac1{2(x+1)^2}(1-x)e^{\frac1{2x+2}}$$

so l'Hospital yields:

$$\lim_{x\to\infty}\left(-e^{\frac1{2x+2}}-\frac1{2(x+1)^2}(1-x)e^{\frac1{2x+2}}\right)=-1-0=-1$$

• @Henry Thank you, you're right...but in fact it is a $\;\frac12\;$ factor, I believe: $$-\frac2{(2x+2)^2}=\frac1{2(x+1)^2}$$ Edited. Jun 15 '16 at 20:43
• Indeed, which is why I said $\frac12 \times 0 =0$ Jun 15 '16 at 20:53

$\lim_\limits{x\to \infty }\frac{\left(\left(1-x\right)e^{\frac{1}{2x+2}}\right)}{x}\\ \lim_\limits{x\to \infty }\frac{1-x}{x}\lim_\limits{x\to \infty }e^{\frac{1}{2x+2}}\\$

• Much simpler, shorter and eleganter (?) . Very nice. +1 Jun 15 '16 at 20:45

$\frac{d}{dx}\left(\left(1-x\right)e^{\dfrac{1}{2x+2}}\right)=-e^{(2x+2)^{-1}}-\dfrac{2(1-x)e^{(2x+2)^{-1}}}{(2x+2)^{2}}$ and letting $x$ go to infinity gives you $-1$