Answer
\[56\]
Work Step by Step
Given that :- $f'$ is continuous and $f(2)=0\;,\; f'(2)=7$
Let \[l=\lim_{x\rightarrow 0}\frac{f(2+3x)+f(2+5x)}{x}\]
Which is $\frac{0}{0}$ form as $f(2)=0$
Using L' Hopitals rule
\[l=\lim_{x\rightarrow 0}\frac{\{f(2+3x)+f(2+5x)\}'}{(x)'}\]
\[l=\lim_{x\rightarrow 0}\frac{3f'(2+3x)+5f'(2+5x)}{1}\]
\[\Rightarrow l=3f'(2)+5f'(2)=8f'(2)=8(7)=56\]
Hence, \[\lim_{x\rightarrow 0}\frac{f(2+3x)+f(2+5x)}{x}=56\]
