Question

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

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