Question

lf $f\left(x\right)=x^2+x+1$, then $\underset{x\to 1}{lim}\frac{f\left(x\right)-f\left(1\right)}{x-1}=$

• A. $3$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

Answer: (A)

$3$

$\underset{x\to 1}{lim}\frac{f\left(x\right)-f\left(1\right)}{x-1}$
This expression is equivalent to $f^{\prime }\left(x\right)$ at $x=1$
Since,
$f\left(x\right)=x^2+x+1$
$\Rightarrow f^{\prime }\left(x\right)=2x+1$
At $x=1$
$f^{\prime }\left(1\right)=2\left(1\right)+1=3$