What is the value of $\lim _ { x \rightarrow - 3} \frac { x ^ { 2} - 9} { x ^ { 2} - 2x - 15}$ ?

Question

27207 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-13T21:16:30

$3/4$

$lim_(x to-3)(x^2-9)/(x^2-2x-15)$

By factoring out the numerator and the denominator,

$=lim_(x to -3)(cancel((x+3))(x-3))/(cancel((x+3))(x-5)) =(-3-3)/(-3-5)=(-6)/(-8)=3/4$

I hope that this was clear.

Answer 2 · 2017-04-13T22:04:05

$lim_(x->-3)(x^2-9)/(x^2-2x-15)=3/4$

Using L'Hopital's rule, we can work out the limit of an expression.

L'Hopital's rule:

$lim_(x->a)(f(x))/(g(x))=lim_(x->a)(f'(x))/(g'(x))$

$lim_(x->-3)(x^2-9)/(x^2-2x-15)=lim_(x->-3)(x^2-9)^'/(x^2-2x-15)^'$

$=lim_(x->-3)(2x)/(2x-2)$

$=lim_(x->-3)x/(x-1)$

$lim_(x->-3)x/(x-1)=(-3)/(-3-1)=3/4$

What is the value of \lim _ { x \rightarrow - 3} \frac { x ^ { 2} - 9} { x ^ { 2} - 2x - 15} \lim _ { x \rightarrow - 3} \frac { x ^ { 2} - 9} { x ^ { 2} - 2x - 15} ?