Limits practice question 17

Treat $a$ as a fixed number and evaluate the limit as usual: the result will depend on $a$. Which technique handles a square root minus a number, over $x$?

Multiply the top and the bottom by the conjugate $\sqrt{ax + 25} + 5$. The numerator becomes $ax$.

After cancelling $x$, the limit is $\dfrac{a}{\sqrt{25} + 5}$. Set that equal to $\tfrac{3}{2}$.