L'Hopital Rule and limits: [f(2+3x)+f(2+5x)]/x if f(2)=0,...

[grapz]

I am stuck on this question, can you guys give me some help.

if f' is continuous, f( 2 ) = 0, and f ' (2) = 7, evaluate

lim x --> 0 [ f( 2 + 3x ) + f ( 2 + 5x ) ] / x

 

[soroban]

Re: L'Hopital Rule and limits

Hello, grapz!

$\displaystyle \text{If }f'{x}\text{ is continuous, }f( 2 ) = 0\text{ and }f ' (2) = 7$,

. . $\displaystyle \text{evaluate: }\;\lim_{x\to0}\,\frac{f( 2\! +\! 3x ) + f ( 2 + 5x )}{x}$

$\displaystyle \text{Since the limit goes to }\frac{0}{0}\text{, we can apply L'Hopital's Rule.}$

$\displaystyle \lim_{x\to0}\,\left[\frac{3\!\cdot\!f'(2\!+\!3x) + 5\!\cdot\!f'(2\!+\!5x)}{1}\right] \;\;=\;\;3\!\cdot\!f'(2) + 5\!\cdot\!f'(2) \;\;=\;\;3(7) + 5(7) \;\;=\;\;56$