# using definition of derivative

$f(x)=x^3-6x^2+9x-5$ is given.

What is the value of $$\lim_{h\to0}\frac{[f'(1+2h)+f'(3-3h)]}{2h}$$

I tried to use the definition of derivative,and here it seems like the expression will be equal to something like the 2nd derivative of $f(x)$ but I'm confused with $2h$ and $-3h$.

• The answer is given as -15. – guest Aug 22 '13 at 0:13
• you mean confused :) – nbubis Aug 22 '13 at 0:14
• Thanks :) It should be mathandgrammer.stackexchange :) – guest Aug 22 '13 at 0:15
• Do you know l'Hopital's rule? – Ataraxia Aug 22 '13 at 0:19
• Yes but I use it when the limit results in 0/0 or inf/inf. Can we use it for this question? – guest Aug 22 '13 at 0:21

Hint: Note that $f'(1)=f'(3)=0$. Rewrite our limit as $$\lim_{h\to 0} \frac{f'(1+2h)-f'(1)}{2h}+\lim_{h\to 0}\left(-\frac{3}{2}\right)\frac{f'(3-3h)-f'(3)}{-3h}.$$

• I was looking for such an answer but I didnt see that f'(1)=f'(3)=0. I added -f'(1)+f'(1) [ and the same for f'(3) ] so they made the expression a bit complicated for me.Thank you – guest Aug 22 '13 at 0:36

We have $$f'(x)=3x^2-12x+9$$ and notice that $$f'(1)=f'(3)=0$$ Now by the definition of the derivative we have $$\lim_{h\to0}\frac{[f'(1+2h)+f'(3-3h)]}{2h}=\lim_{u\to0}\frac{f'(1+u)-f'(1)}{u}-\frac{3}{2}\lim_{v\to0}\frac{f'(3+v)-f'(3)}{v}\\ =f''(1)-\frac{3}{2}f''(3)=-15$$

Hints:

• $f'(1+ 2 h) = 9 - 12 (1 + 2 h) + 3 (1 + 2 h)^2$
• $f'(3-3h) =9-12 (3-3 h)+3 (3-3 h)^2$
• $f'(1+2h) + f'(3-3h) = 8-12 (3-3 h)+3 (3-3 h)^2-12 (1+2 h)+3 (1+2 h)^2$

So, we have:

$$\lim_{h\to0}\frac{[f'(1+2h)+f'(3-3h)]}{2h} = \lim_{h\to0}\dfrac{-30 h + 39 h^2}{2h} = \lim_{h\to0}\dfrac{-30 + 39 h}{2} = -15$$ Can you fill in the details?

This is not the second derivative, since the points the first derivative is evaluated at are not the same points. That being said, you can just calculate the first derivative, plug in $1+2h, 3-3h$, and see what you get.

For reference, note that you'll get (after some simplification): $$\lim_{h\to 0} \frac{3h(13h-10)}{2h}=\lim_{h\to 0} \frac{-30h}{2h}=-15$$