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  4. If f(3) = 4 and f'(3) = 1, then displaystyle limx → 3 (xf(3)-3 f(x)/x-3)

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Q. If $f(3) = 4$ and $f'(3) = 1$, then $\displaystyle \lim_{x \to 3}$ $\frac{xf\left(3\right)-3\,f\left(x\right)}{x-3}$

Limits and Derivatives

Solution:

We have, $\displaystyle \lim_{x \to 3}$ $\frac{xf\left(3\right)-3\,f\left(x\right)}{x-3}$
$=\displaystyle \lim_{x \to 3}$ $\frac{\left(x-3\right)f\left(3\right)+3\left\{f\left(3\right)-f\left(x\right)\right\}}{x-3}$
$=\displaystyle \lim_{x \to 3}$$\frac{\left(x-3\right)f\left(3\right)}{x-3}$$+\displaystyle \lim_{x \to 3}$$\frac{-3\left\{f\left(x\right)-f\left(3\right)\right\}}{x-3}$
$=f\left(3\right)+\left(-3\right)$$\displaystyle \lim_{x \to 3}$ $\frac{f\left(x\right)-x\left(3\right)}{x-3}$
$=f\left(3\right)-3f'\left(3\right)$
$=4-3\times1=1$.