Question

Evaluate the limit, if it exists. Show work. lim┬(x→5)⁡〖(x^2-3x-10)/(2x-10)〗

Answer 1

Hello, first of all we can notice that

$5^2-3*5-10=25-15-10=0$

and

$2*5-10=0$

and 0 divided by 0 is undetermined.

So, we need to factorise and simplify first.

$x^2-3x-10=(x-5)(x+2)\\\\\text{Because we already known that 5 is a zero...}\\\\\text{... and the sum of the zeroes is 3 = 5 - 2 and the product is -10 = 5 * (-2).}$

For x different from 5, we can write

$\dfrac{x^2-3x-10}{2x-10}=\dfrac{(x-5)(x+2)}{2(x-5)}=\dfrac{x+2}{2}$

So, the limit when x tends to 5 is

$\dfrac{5+2}{2}=\boxed{\dfrac{7}{2}}$

Thank you