• DLS

Limits question [Challenge]

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.

  • DLS

Limits question [Challenge]

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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

  • DLS
\[\LARGE \lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{x} -e}{x}\]
* :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

  • DLS
first time in life :O
Taylor Series expansion?
  • DLS
Use whatever you want.
Ans 0?
  • DLS
Nope
Taylor series works
Weird. The Taylor Series expansion is in the terms of \(e\) and \(x\). The first term is \(e\)
  • DLS
Taylor series gives 0?
gives -e/2 if you calculate correctly
Oh wait! lol, yeah. I forgot to consider that -e/2 which had no \(x\)
  • DLS
How about a different way rather than taylor series?
Definition of derivative? :-|
  • DLS
o.O
I was just thinking how this might be the definition of a derivative in disguise.
Binomial expansion might work ... but that's too long
DLS Use whatever you want. 15 minutes ago DLS How about a different way rather than taylor series? 9 minutes ago
  • DLS
You are right,still investigating for more methods,I wrote so to specify that you are not bound to any particular method,just asking if you know anything alternate :)
L'hopital works
http://www.wolframalpha.com/input/?i=Limit%5B%281+%2B+x%29%5E%281%2Fx%29+%281%2F%28x+%281+%2B+x%29%29+-+Log%5B1+%2B+x%5D%2Fx%5E2%29%2C+x+-%3E+0%5D after using L'hopital you get this expression. The first term tends to e .. if you again apply L'hopital on the second term, you will get -1/2

Not the answer you are looking for?

Search for more explanations.

Ask your own question