Question

How would I evaluate the following limit:

Answer 1

\[\Huge Evaluate:\lim_{x \rightarrow 1} \frac {x^3-1}{x^2-1}?\]

Answer 2

\[\lim_{x \rightarrow 1} \frac {x^3-1}{x^2-1}\]\[ ^{L'H}=\lim_{x \rightarrow 1} \frac {3x^2}{2x}\]\[ ^{L'H}=\lim_{x \rightarrow 1} \frac {6x}{2}\]\[=\lim_{x\rightarrow 1}3x\]\[=\]

Answer 3

Where does the three and two come from?

Answer 4

I would try factoring the numerator and the denominator using these little formulas and then cancelling common expression in the numerator / denominator:
a^2 - b^2 = (a + b)(a - b)
a^3 - b^3 = (a - b)(a^2 + ab + b^2

Answer 5

\[(x^3 - b^3)= (x-b)(x^2 + bx + b^2)\]
\[(x^2 - b^2)= (x-b)(x+b)\]
So how about you factor the top and bottom

Answer 6

L'Hôpital's rule
when the limit of a fraction has infinities or zeros you can differentiate both the numerator and denominator and the limit of this will be equal to the true measure of the limit

Answer 7

I'm in Honors PreCalculus...

Answer 8

Ok then just factor as we told you if you haven't learnt L'Hôpital's rule

Answer 9

\[\lim_{x \rightarrow 1}((x-1)(x^2 +x+1^2))/((x+1)(x-1))\]

Answer 10

Can you go on now?

Answer 11

Yes, it makes sense now.

Answer 12

Great.

Answer 13

L'Hôpital's rule can be applied if

\[\lim_{x\rightarrow c}={f(x)}=\lim_{x\rightarrow c}={g(x)}=0 \text{ or } \pm \infty\]
\[\lim_{x\rightarrow c}=\frac{f(x)}{g(x)}\qquad^{L'H}=\lim_{x\rightarrow c}=\frac{f'(x)}{g'(x)}\]

Answer 14

Thank You, @UnkleRhaukus & @AccessDenied !

Answer 15

And @beeqay !

Answer 16

I'm not sure you could use L'Hopital's rule that second time since x->1 of that function was already in form 3x^2 / x, and x->1 didn't result in any issues.

You're welcome. :)

Answer 17

typo**
\[\lim_{x\rightarrow c}{f(x)}=\lim_{x\rightarrow c}{g(x)}=0 \text{ or } \pm \infty\]

Answer 18

No problem.