Search any question & find its solution
Question:
Answered & Verified by Expert
If $f: R \rightarrow R$ is such that $f(3)=16, f^{\prime}(3)=4$, then $\lim _{x \rightarrow 3} \frac{x f(3)-3 f(x)}{x-3}=$
Options:
Solution:
1310 Upvotes
Verified Answer
The correct answer is:
4
$f(3)=16 f^{\prime}(3)=4$
$\Rightarrow \quad \lim _{x \rightarrow 3} \frac{x f(3)-3 f(x)}{x-3} \quad\left[\frac{0}{0}\right.$ form $]$
By $L^{\prime}$ Hospital Rule, $\lim _{x \rightarrow 3} \frac{f(3)-3 f^{\prime}(x)}{1}$
$\Rightarrow f(3)-3 f^{\prime}(3) \quad \Rightarrow 16-3(4) \Rightarrow 4$
$\Rightarrow \quad \lim _{x \rightarrow 3} \frac{x f(3)-3 f(x)}{x-3} \quad\left[\frac{0}{0}\right.$ form $]$
By $L^{\prime}$ Hospital Rule, $\lim _{x \rightarrow 3} \frac{f(3)-3 f^{\prime}(x)}{1}$
$\Rightarrow f(3)-3 f^{\prime}(3) \quad \Rightarrow 16-3(4) \Rightarrow 4$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.