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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[\lim_{x\to\infty} (x+x^{-1})e^{1/x}-x\]?
Ya this is the question. what's the solution?
Hint : as x approaches infinity 1/x approaches zero so e^1/x will approach 1.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\[\lim_{x\to\infty} (x+x^{-1})e^{1/x}-x \\~\\= \lim_{x\to\infty} xe^{1/x}+x^{-1}e^{1/x}-x\\~\\ = \lim_{x\to\infty} x(e^{1/x}-1)+x^{-1}e^{1/x}\] 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.
@Pawanyadav Does that help?
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.
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.
|dw:1433801268065:dw|
The answer After solving is 1.
The answer After solving is 1.
The answer After solving is 1.
The answer After solving is 1.
@Zarkon I think something is buggy.
I see so many replies "The answer After solving is 1."

Not the answer you are looking for?

Search for more explanations.

Ask your own question