G grapz Junior Member Joined Jan 13, 2007 Messages 80 Dec 3, 2007 #1 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
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
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Dec 3, 2007 #2 Re: L'Hopital Rule and limits Hello, grapz! \(\displaystyle \text{If }f'{x}\text{ is continuous, }f( 2 ) = 0\text{ and }f ' (2) = 7\), . . \(\displaystyle \text{evaluate: }\;\lim_{x\to0}\,\frac{f( 2\! +\! 3x ) + f ( 2 + 5x )}{x}\) Click to expand... \(\displaystyle \text{Since the limit goes to }\frac{0}{0}\text{, we can apply L'Hopital's Rule.}\) \(\displaystyle \lim_{x\to0}\,\left[\frac{3\!\cdot\!f'(2\!+\!3x) + 5\!\cdot\!f'(2\!+\!5x)}{1}\right] \;\;=\;\;3\!\cdot\!f'(2) + 5\!\cdot\!f'(2) \;\;=\;\;3(7) + 5(7) \;\;=\;\;56\)
Re: L'Hopital Rule and limits Hello, grapz! \(\displaystyle \text{If }f'{x}\text{ is continuous, }f( 2 ) = 0\text{ and }f ' (2) = 7\), . . \(\displaystyle \text{evaluate: }\;\lim_{x\to0}\,\frac{f( 2\! +\! 3x ) + f ( 2 + 5x )}{x}\) Click to expand... \(\displaystyle \text{Since the limit goes to }\frac{0}{0}\text{, we can apply L'Hopital's Rule.}\) \(\displaystyle \lim_{x\to0}\,\left[\frac{3\!\cdot\!f'(2\!+\!3x) + 5\!\cdot\!f'(2\!+\!5x)}{1}\right] \;\;=\;\;3\!\cdot\!f'(2) + 5\!\cdot\!f'(2) \;\;=\;\;3(7) + 5(7) \;\;=\;\;56\)
