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.

See more answers at brainly.com

The general solution amounts to applying l'Hopital's rule. Are you familiar?

No Im not. Can you show me?

Im sorry D: Im still a bit confused :/

Okay wait let me try continuing on paper. That explanation helped a bit.

\[\lim_{x\to0}\frac{e^x-1}{x}\to\lim_{x\to0}\frac{\frac{d}{dx}(e^x-1)}{\frac{d}{dx}x}\]

brb quick lunch. @Ishaan94 @eliassaab help here please

Amanderp? Is it still confusing you?

YES :(

What part is confusing you?

So I guess the rule itself, including the application is confusing me.

Im sorry. I need a refresh. Differentiation is simply taking the derivative right?

Yes.

Yes.

Like it wouldnt be evaluated, so to speak, with the 0. It would be left in terms of the variables?

I am sorry. I am not really a good teacher. I hope @eliassaab can help you.

Lol I appreciate your honesty.
Im not a math person so its just not clicking :3

\[\sin x \approx x\]For lower values of x. Maybe that's why \[\lim_{x\to0} \frac{\sin x}x = 1\]

See also this
http://en.wikipedia.org/wiki/Differentiation_of_trigonometric_functions

Interesting. Thanks for the correction. :)