Ap calculus: lim, x ->0, of [ x (e^x + 1/x) ]

[joy08]

Hello. I'm having trouble with this problem:
Find the limit x (e^x+1/x)
x->0
This is what I did:
Lim 0 (e^o+1/0)
0x1=0+1/0=0
so the limit does not exist

 

[o_O]

Re: Ap calculus

Mmm, what you get at the end is an indeterminant form, essentially in the form of (0 x ?). I don't know if you haven't covered this yet but basically, there are certain limits that can't be evaluated in a given form usually involving 0's and ?'s. For this problem, try multiplying x through.

\(\displaystyle \lim_{x \to 0} \left[x \left(e^{x} + \frac{1}{x}\right)\right] = \:\:???\)

 

[galactus]

I'm sorry. I misread your post before.

I assume your limit problem is \(\displaystyle \lim_{x\to{0}}x(e^{x}+\frac{1}{x})\)

If you distibute you get:

\(\displaystyle \lim_{x\to{0}}xe^{x}+\lim_{x\to{0}}1\)

Now, can you see what the limit is?. It does have a limiting value and is not undefined.