Question

If $f\left(x\right)=x^2-3x+5$, then $\underset{x\to 2}{lim}\frac{f\left(x\right)-f\left(2\right)}{x-2}$

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. $4$

Answer 1

The correct option is B $1$

The above given expression,
${lim}_{x\to 2}\frac{f\left(x\right)-f\left(2\right)}{x-2}$ is equivalent to $f^{\prime }\left(x\right)$ at $x=2$

$f\left(x\right)=x^2-3x+5$

$\Rightarrow f^{\prime }\left(x\right)=2x-3$

$\Rightarrow f^{\prime }\left(2\right)=2\left(2\right)-3$

$=1$