Question

How to solve this?
lim x approache 1 (xe^x-e^x)/x^2-1

Answer 1

Factor the bottom as difference of squares.
Factor the numerator.
Something will cancel.

Answer 2

\[ \lim_{x \rightarrow 1} (xe ^{x}-e ^{x})\div[(x ^{2})-1]\]

Answer 3

though L'hospital's rule is harder to use here than my method

Answer 4

yeah, it's because I wrote e^1/x instead of e^1/2, but the result is right

Answer 5

I would not write the 1, just e/2

Answer 6

\[\lim_{x \rightarrow 1}[xe^x+e^x−e^x]/2x = \lim_{x \rightarrow 1}xe^x/2x = \lim_{x \rightarrow 1}e^x/2 = e^1/2=e/2\]

Answer 7

So there you have it, two ways to solve this problem :)

Answer 8

done

Answer 9

the answer must be e/2. did you get it? Thanks --- Terima kasih(in malay)

Answer 10

Yes, you can get that by either factoring, or using L'hospital's rule. Leebaka has done all the work for the L'hospital method.

Answer 11

Sorry.. I still do not understand that solution Leebaka.. Please help me :-(

Answer 12

What do you still don't understand?

Answer 13

Do you know L'hospital's rule?

Answer 14

I just derived the top and bottom of the function and then substituted the x for 1

Answer 15

if L'hospital's rule is confusing you try my way:
Factor the numerator and denominator, look for something to cancel (it will)

Answer 16

I did not know L'hospital rule. Can you give that rule?

Answer 17

Just use Lim f'(x)/g'(x) instead of Lim f(x)/g(x). You just need to use the derivative of the numerator and denominator, in more complex cases it is more useful than using factoration.

Answer 18

If you haven't done it using factoration yet:
\[\lim_{x \rightarrow 1}(xe^x-e^x)/(x^2-1) = \lim_{x \rightarrow 1} e^x(x-1)/(x+1)(x-1) = \lim_{x \rightarrow 1}e^x/(x+1) = e^1/(1+1) = e/2\]