To find the limit

$\underset{x5}{lim}\frac{2x^2-50}{3x-15}$

we start by substituting $x=5$ :

$\frac{2(5{)}^2-50}{3(5)-15}=\frac{2(25)-50}{15-15}=\frac{50-50}{0}=\frac{0}{0}.$

This is an indeterminate form, which means we cannot directly evaluate the limit$.$ To resolve this, we will factor both the numerator and the denominator.

Factor the numerator: The expression $2x^2-50$ can be factored as follows:

$2x^2-50=2(x^2-25)=2(x-5)(x5).$

Factor the denominator: The expression $3x-15$ can be factored as:

$3x-15=3(x-5)$

Now, we substitute these factored forms into our limit:

$\frac{2x^2-50}{3x-15}=\frac{2(x-5)(x5)}{3(x-5)}$

Cancel the common factor: The $(x-5)$ term appears in both the numerator and the denominator. We can cancel it $($as long as $x5)$:

$\frac{2(x5)}{3}$

Evaluate the limit: Now we can find the limit as $x$ approaches $5$ :

$\underset{x5}{lim}\frac{2(x5)}{3}=\frac{2(55)}{3}=\frac{2(10)}{3}=\frac{20}{3}$

Thus, the final answer to the limit is:

$\frac{20}{3}.$