Question

Solve: $\underset{x\to \frac{1}{2}}{lim}\frac{4x^2+8x-5}{1-4x^2}$.

• A. $-3$
• B. $-\frac{1}{2}$
• C. $0$
• D. $\frac{1}{2}$
• E. $3$

Answer 1

The correct option is A $-3$

When $x$ tends to $\frac{1}{2}$, denominator tends to $0$ and numerator tends to $0$.

Therefore $\frac{1}{2}$ is factor of both numerator and denominator.

So first simplify it, we get
$\frac{4x^2+8x-5}{1-4x^2}=\frac{(2x+5)(2x-1)}{(1-2x)(1+2x)}$

$=\frac{-(2x+5)}{(2x+1)}$

Now put $x=\frac{1}{2}$, we get
$\frac{-\left(2\right(1/2)+5)}{2\left(1/2\right)+1}=\frac{-6}{2}=-3$