Limits

[enzzza]

I need help i am trying to graph function but i have problem with this type of limits

i found vertical asymptote and all other steps.
I tried to bring(x+1) down to denominator tried same with e but cant set up everything for L'Hospital's Rule. Can some one try to explain me what is algorithm to solving this type of limits.

 

[Limax]

Hint:

\(\displaystyle \displaystyle \ \lim_{x\to\infty}(x+1)e^{\frac1{2x}}-x\ =\ \lim_{x\to0^+}\left(\frac1x+1\right)e^{\frac x2}-\frac1x\ =\ \lim_{x\to0^+}\frac{(1+x)e^{\frac x2}-1}x\)​

Now you can use L’Hôpital.

 

[enzzza]

Hint:

\(\displaystyle \displaystyle \ \lim_{x\to\infty}(x+1)e^{\frac1{2x}}-x\ =\ \lim_{x\to0^+}\left(\frac1x+1\right)e^{\frac x2}-\frac1x\ =\ \lim_{x\to0^+}\frac{(1+x)e^{\frac x2}-1}x\)​

Now you can use L’Hôpital.

Thank you so much but i rlly don't understand how is posible to change infinty to 0 :S i wish i could . Are there some rules that are allowed here