Question

Find the value of the limit: $\underset{x\to 1}{lim}\frac{x^2-6x+5}{x^2+3x-4}$

• A. $-1.25$
• B. $-0.80$
• C. $0.80$
• D. $1.25$
• E. The limit does not exist

Answer 1

The correct option is B $-0.80$

At $x=1$, both numerator and denominator is becoming zero.

So, we should factorize first before solving. we get
$\frac{x^2-6x+5}{x^2+3x-4}=\frac{(x-1)(x-5)}{(x-1)(x+4)}=\frac{x-5}{x+4}$
Now substitute $x=1$ , we get

$\frac{x-5}{x+4}=\frac{1-5}{1+4}$

$=\frac{-4}{5}$

$=-0.8$