The problem is to find the derivative of $f(x) = \frac{3x}{x^2+1}$ at $x = -4$ using the limit definition, $$ f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} $$


I plug in $-4$ for $x$ when using the limit definition and I always end up stuck with an unfactorable denominator. I tried it $5$ times already but I always end up with the same denominator.

$$\frac{-45 + 12h}{17 (h^2 - 8h + 17)}$$

  • $\begingroup$ When you have that at the end, you can safely plug in $h=0$, because it's not an indeterminate form (i.e. it's not $\dfrac00$). $\endgroup$ – Akiva Weinberger Sep 3 '14 at 1:21

If you want to find $$\lim_{h\rightarrow 0}\frac{-45 + 12h}{17 (h^2 - 8h + 17)}$$ as $h$ approaches $0$ the you can see that it becomes (since the numerator and denominator are such that when $h$ is $0$ we don't end up with and indeterminate form): $$ -\frac{45}{17\times17} =-\frac{45}{289}$$ Which if you work out the derivative normally you can see that this is the case since: $$ f'(x)=-\frac{3 \left(x^2-1\right)}{\left(x^2+1\right)^2} \;\; \therefore \;\; f'(4)=-\frac{45}{289}$$

Just as an extension to the above you can see that in general you have: $$f'(x)=\frac{\frac{3 (h+x)}{(h+x)^2+1}-\frac{3 x}{x^2+1}}{h} \;\;\; \text{which can simplify to}\;\; -\frac{3 (x (h+x)-1)}{\left(x^2+1\right) \left((h+x)^2+1\right)} $$ which again if you let $h=0$ then you can see that it becomes: $$f'(x)=-\frac{3 (x (0+x)-1)}{\left(x^2+1\right) \left((0+x)^2+1\right)}=-\frac{3 (x^2-1)}{\left(x^2+1\right) \left(x^2+1\right)}=-\frac{3 \left(x^2-1\right)}{\left(x^2+1\right)^2}$$


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