Hello, first of all we can notice that
[tex]5^2-3*5-10=25-15-10=0[/tex]
and
[tex]2*5-10=0[/tex]
and 0 divided by 0 is undetermined.
So, we need to factorise and simplify first.
[tex]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).}[/tex]
For x different from 5, we can write
[tex]\dfrac{x^2-3x-10}{2x-10}=\dfrac{(x-5)(x+2)}{2(x-5)}=\dfrac{x+2}{2}[/tex]
So, the limit when x tends to 5 is
[tex]\dfrac{5+2}{2}=\boxed{\dfrac{7}{2}}[/tex]
Thank you
