From the definition, we have

With
, we get
. So the limit we want to compute is

Rationalize the numerator by multiplying by its conjugate:

x is approaching 10, which is to say x ≠ 10, so we can cancel the factors of x - 10 and remove the discontinuity.
Then we're left with

evaluated by direct substitution, which we can do since the limand is continuous.