Question

Find $\underset{t\to 4}{lim}\frac{t-\sqrt{3t+4}}{4-t}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

If we substitute t=4, we get $\frac{0}{0}$ form

So, We will rationalize the given expression. (We will multiply both denominator and numerator by $t+\sqrt{3t+4}$.

$\underset{t\to 4}{lim}\frac{t-\sqrt{3t+4}}{4-t}=\underset{t\to 4}{lim}\frac{t-\sqrt{3t+4}}{4-t}\frac{(t+\sqrt{3t+4})}{(t+\sqrt{3t+4})}$

$\underset{t\to 4}{lim}\frac{t^2-3t-4}{(4-t)(t+\sqrt{3t+4})}$

$\underset{t\to 4}{lim}\frac{(t-4)(t+1)}{(4-t)(t+\sqrt{3t+4})}$

$\underset{t\to 4}{lim}\frac{-(t+1)}{t+\sqrt{3t+4}}$

$=\frac{-5}{4+\sqrt{16}}$

$=\frac{-5}{8}$