Question

It is given that $f^{\prime }(a)$ exists, then $\underset{x\to a}{\lim }$ $\frac{xf\left(a\right)-af\left(x\right)}{(x-a)}$ is equal to

• A. $f(a)-a{f}^{\prime }(a)$
• B. $f^{\prime }(a)$
• C. $-f^{\prime }(a)$
• D. $f(a)+a{f}^{\prime }(a)$

Answer 1

Answer: (A)

$\underset{x\to a}{\lim }$ $\frac{xf\left(a\right)-af\left(x\right)}{x-a}$
$=\underset{x\to a}{\lim }$ $\frac{xf\left(a\right)-xf\left(x\right)+xf\left(x\right)-af\left(x\right)}{\left(x-a\right)}$
$=\underset{x\to a}{\lim }$ $\left[\frac{x\left[f\left(a\right)-f\left(x\right)\right]}{\left(x-a\right)}+\frac{f\left(x\right)\left(x-a\right)}{\left(x-a\right)}\right]$
$=\underset{x\to a}{\lim }$ $-\frac{x\left[f\left(x\right)-f\left(a\right)\right]}{x-a}+$$\underset{x\to a}{\lim }f(x)$
$=-a{f}^{\prime }(a)+f(a)$
$=f(a)-a{f}^{\prime }(a)$