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ankara55t · Oct 24, 2009, 02:41 PM

How can I use lim as h approaches 0 in evaluating

f(x) = 5 + h/x + 7 using f(x) = [f(x + h) -f(x)]/h ?

galactus · Oct 24, 2009, 05:42 PM

Please be sure your problem statement is correct. Should that perhaps be

$\frac{5+x}{x+7}$ ?

$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ is the definition of a derivative.

$\lim_{h\to 0}\left[\frac{\frac{5+x+h}{x+h+7}-\frac{5+x}{x+7}}{h}\right]$

Cross multiply the numerator and get:

$\lim_{h\to 0}\left[\frac{(5+x+h)(x+7)-(5+x)(x+h+7)}{h(x+h+7)(x+7)}\right]$

All the terms in the numerator cancel except 2h and we are left with:

$\lim_{h\to 0}\frac{2h}{h(x+h+7)(x+7)}$

$=\frac{2}{(x+7)^{2}}$

which is the derivative of $\frac{x+5}{x+7}$

See how it works now?

Also, if you are in a calculus class you should know how to use grouping symbols correctly.

The way you have it written means $5+\frac{h}{x}+7$

Is it safe to assume that is not what you meant?

If $\frac{5+h}{x+7}$ is correct, then let h=0.

Perhaps the original problem was to find the derivative of $5ln(x+7)$ using first principles?

This derivative is $\frac{5}{x+7}$ .