Math Help - Difference Quotient

1. Difference Quotient

I have a problem in which i have to solve for a limit:

$lim$ $f(2+h) - f(2)/h$
$h\rightarrow 0$

$f(x) = 2 / x-1$

The divide sign is divided by the whole thing not just f(2) and that goes for f(x) also

2. The limit is $f'(2)$ you need to differeniate $f(x)$ and plug in the appropriate value.

Bobak

3. Originally Posted by snakeman11689
I have a problem in which i have to solve for a limit:

$lim$ $f(2+h) - f(2)/h$
$h\rightarrow 0$

$f(x) = 2 / x-1$

The divide sign is divided by the whole thing not just f(2) and that goes for f(x) also
$f(x)=\frac{2}{x-1}$?

If so

$\lim_{\Delta{x}\to{0}}\frac{\frac{2}{x+\Delta{x}-1}-\frac{2}{x-1}}{\Delta{x}}$

Combining fractions we get

$\lim_{\Delta{x}\to{0}}\frac{\frac{-2\Delta{x}}{x^2+x(\Delta{x})-2x-\Delta{x}+1}}{\Delta{x}}=\lim_{\Delta{x}\to{0}}\fr ac{-2}{x^2+x(\Delta{x})-2x-\Delta{x}+1}$

And direct substitution yields

$\frac{-2}{x^2-2x+1}=\frac{-2}{(x-1)^2}$

4. Originally Posted by Mathstud28
$f(x)=\frac{2}{x-1}$?

If so

$\lim_{\Delta{x}\to{0}}\frac{\frac{2}{x+\Delta{x}-1}-\frac{2}{x-1}}{\Delta{x}}$

Combining fractions we get

$\lim_{\Delta{x}\to{0}}\frac{\frac{-2\Delta{x}}{x^2+x(\Delta{x})-2x-\Delta{x}+1}}{\Delta{x}}=\lim_{\Delta{x}\to{0}}\fr ac{-2}{x^2+x(\Delta{x})-2x-\Delta{x}+1}$

And direct substitution yields

$\frac{-2}{x^2-2x+1}=\frac{-2}{(x-1)^2}$
hmm, the answer is -2 rite???

5. Originally Posted by snakeman11689
hmm, the answer is -2 rite???
$\lim_{\Delta{x}\to{0}}\frac{f(x+\Delta{x})-f(x)}{\Delta{x}}=f'(x)\Rightarrow{\lim_{\Delta{x}\ to{0}}\frac{f(2+\Delta{x})-f(2)}{\Delta{x}}=f'(2)}$

So letting $f(x)=\frac{2}{x-1}$

we have as shown above $f'(x)=\frac{-2}{(x-1)^2}$

$\therefore\lim_{\Delta{x}\to{0}}\frac{f(2+\Delta{x })-f(2)}{\Delta{x}}=f'(2)=\frac{-2}{(2-1)^2}=-2$

6. Originally Posted by Mathstud28
$\lim_{\Delta{x}\to{0}}\frac{f(x+\Delta{x})-f(x)}{\Delta{x}}=f'(x)\Rightarrow{\lim_{\Delta{x}\ to{0}}\frac{f(2+\Delta{x})-f(2)}{\Delta{x}}=f'(2)}$

So letting $f(x)=\frac{2}{x-1}$

we have as shown above $f'(x)=\frac{-2}{(x-1)^2}$

$\therefore\lim_{\Delta{x}\to{0}}\frac{f(2+\Delta{x })-f(2)}{\Delta{x}}=f'(2)=\frac{-2}{(2-1)^2}=-2$

got it , thank you so much for helping me