## Pawanyadav one year ago Lim{(x+x^-1)e^(1÷x)-x=? Where x tending to infinity.

1. geerky42

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

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

3. welshfella

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

4. geerky42

$\lim_{x\to\infty} (x+x^{-1})e^{1/x}-x \\~\\= \lim_{x\to\infty} xe^{1/x}+x^{-1}e^{1/x}-x\\~\\ = \lim_{x\to\infty} x(e^{1/x}-1)+x^{-1}e^{1/x}$ 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.

5. geerky42

@Pawanyadav Does that help?

6. anonymous

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

Yes absolutly

The answer After solving is 1.

The answer After solving is 1.

The answer After solving is 1.

11. geerky42

Correct.

The answer After solving is 1.

The answer After solving is 1.

The answer After solving is 1.

The answer After solving is 1.

The answer After solving is 1.

The answer After solving is 1.

18. geerky42

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The answer After solving is 1.

The answer After solving is 1.

The answer After solving is 1.

The answer After solving is 1.

23. geerky42

@Zarkon I think something is buggy.

24. geerky42

I see so many replies "The answer After solving is 1."