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 ?

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}.