N nikchic5 Junior Member Joined Feb 16, 2006 Messages 106 Mar 21, 2006 #1 If f' is continuous, f(9)=0 and f'(9)=8, evaluate the limit as x approaches 0 for ( f (9 + 7x) + f (9 + 8x) ) / ( x ) Thanks so much for any help!
If f' is continuous, f(9)=0 and f'(9)=8, evaluate the limit as x approaches 0 for ( f (9 + 7x) + f (9 + 8x) ) / ( x ) Thanks so much for any help!
R royhaas Full Member Joined Dec 14, 2005 Messages 832 Mar 21, 2006 #2 Are you supposed to use L'Hospital's Rule? If not, start with f(9+7x)/x = 7( f(9+7x)-f(9))/(7x).
N nikchic5 Junior Member Joined Feb 16, 2006 Messages 106 Mar 21, 2006 #3 Ok... Hey I don't think I need to use L'Hospitals rule....so where do I go from there?
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Mar 21, 2006 #4 \(\displaystyle \L \begin{array}{l} \frac{{f(9 + 7x) + f(9 + 8x)}}{x} = 7\frac{{f(9 + 7x) - f(9)}}{{7x}} + 8\frac{{f(9 + 8x) - f(9)}}{{8x}} \\ lim_{7x \to 0} 7\frac{{f(9 + 7x) - f(9)}}{{7x}} = 7f'(9) \\ \end{array}\). Can you finish it?
\(\displaystyle \L \begin{array}{l} \frac{{f(9 + 7x) + f(9 + 8x)}}{x} = 7\frac{{f(9 + 7x) - f(9)}}{{7x}} + 8\frac{{f(9 + 8x) - f(9)}}{{8x}} \\ lim_{7x \to 0} 7\frac{{f(9 + 7x) - f(9)}}{{7x}} = 7f'(9) \\ \end{array}\). Can you finish it?
N nikchic5 Junior Member Joined Feb 16, 2006 Messages 106 Mar 21, 2006 #5 Ok... Yes...I got the answer as 120; Is that correct? Thanks so much
