i'm so lost on this problem.. I don't even know where to start on this one. Help me guys
Evaluate.
Lim (3+h)^-1 - 3^-1 / h
h->0
$\displaystyle \lim_{h \to 0} \frac{\displaystyle\frac{1}{3+h} - \frac{1}{3}}{h}$
ok like chop suey said find a common denominator
$\displaystyle \lim_{h \to 0}\frac{\frac{-h}{9+3h}}{h}$
$\displaystyle \lim_{h \to 0} \frac{-h}{h(9+3h)}$
Can you finish up from here?
Sorry to bump this thread again but I had another question pertaining to the problem from before.
The slope would equal -1/9 right? My professor had this answer down.
"The slope of the tangent line to y = 1/x at the point (3, 1/3) is -1/9."
How did he get the points (3,1/3) and y = 1/x?
The derivative is the slope of the tangent line at a point x=a. The definition is:
$\displaystyle \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
This is very similar to your limit. The given question had already plugged in the point, but one can find the original function easily. Let's revert 3 back to x:
$\displaystyle f(x+h) = \frac{1}{x+h}$
$\displaystyle f(x) = \frac{1}{x}$
$\displaystyle f(3) = \frac{1}{3}$
So $\displaystyle -\frac{1}{9}$ is the slope of the tangent line to point $\displaystyle (3, \frac{1}{3})$