Lim{(x+x^-1)e^(1÷x)-x=?
Where x tending to infinity.

Question

Lim{(x+x^-1)e^(1÷x)-x=?
Where x tending to infinity.

geerky42 · Answer

$$\lim_{x\to\infty} (x+x^{-1})e^{1/x}-x$$?

pawanyadav · Answer

Ya this is the question. what's the solution?

welshfella · Answer

Hint : as x approaches infinity 1/x approaches zero so e^1/x  will approach 1.

geerky42 · Answer

$$\lim_{x\to\infty} (x+x^{-1})e^{1/x}-x \~\= \lim_{x\to\infty} xe^{1/x}+x^{-1}e^{1/x}-x\~\ = \lim_{x\to\infty} x(e^{1/x}-1)+x^{-1}e^{1/x}$$

Here, $x^{-1}e^{1/x}$ approaches to 0.
So we are left with $$\lim_{x\to\infty} x(e^{1/x}-1)\~\=\lim_{x\to\infty} \dfrac{e^{1/x}-1}{\dfrac{1}{x}}$$From here, use L'Hopital's Rule.

geerky42 · Answer

@Pawanyadav Does that help?

anonymous · Answer

I don't think so because it's not the solution :P

pawanyadav · Answer

Yes  absolutly

pawanyadav · Answer