Finding the value of limit $\lim_{h\to0} \frac{f(a+h)-f(a+3h)}{2h}$

Question

Given that $f^{\prime}(a)=5$ (the function can be defined for $x=a$), find

$$\lim_{h\to0} \frac{f(a+h)-f(a+3h)}{2h} =?$$

I have tried using these two formulas that define the derivatives:

1) $\lim_{h\to0} \frac{f(a+h)-f(a-h)}{2h} =5$

2) $\lim_{h\to0} \frac{f(a+h)-f(a)}{h} =5$

but, seemingly I can't achieve an implicit algebric expression that will help me solve the problem... I'mm really stumped and can't seem to figure out, how to even approach it? (Btw I have not an answer...)

Answer 1

gimusi's (formal) approach may well be preferable to my informal approach, which may be the perspective that the questioner intended.

Anyway, $f'(a) = 5 \;$ implies that as $\;h \rightarrow 0, \;f(a + kh) \rightarrow f(a) + 5kh.$

Therefore, $\dfrac{f(a+h) - f(a+3h)}{2h} \rightarrow \dfrac{5h - 15h}{2h} = -5.$

Answer 2

Note that

$$\frac{f(a+h)-f(a+3h)}{2h}=\frac12\frac{f(a+h)-f(a)}{h}-\frac32\frac{f(a+3h)-f(a)}{3h}$$

then by definition of derivative

$$\lim_{h\to 0}\frac{f(a+h)-f(a+3h)}{2h}=\frac12\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}-\frac32\lim_{3h\to 0}\frac{f(a+3h)-f(a)}{3h}=\\=\frac 125-\frac325=-5$$

or as an alternative by the equivalent definition of differential

$$f(a+h)=f(a)+f'(a)\cdot h+o(h)$$

then

$$\lim_{h\to0} \frac{f(a+h)-f(a+3h)}{2h}=\lim_{h\to0} \frac{f(a)+f'(a)\cdot h-f(a)-f'(a)\cdot3h+o(h)}{2h}=\\=\lim_{h\to0} \frac{-f'(a)\cdot2h+o(h)}{2h}=\lim_{h\to0} \frac{-f'(a)\cdot2+o(1)}{2}=-5$$

Answer 3

In general, if $f$ is differentiable at $a$, you can write $$ f(a+h)=f(a)+hf'(a)+h\varphi(h) $$ where $\lim_{h\to0}\varphi(h)=0$; just set $$ \varphi(h)=\frac{f(a+h)-f(a)}{h}-f'(a) $$ and apply differentiability. Therefore $$ \frac{f(a+h)-f(a+3h)}{2h}= \frac{f(a)+hf'(a)+h\varphi(h)-f(a)-3hf'(a)-3h\varphi(3h)}{2h} $$ and, simplifying, you get $$ -f'(a)+\frac{1}{2}\varphi(h)-\frac{3}{2}\varphi(3h) $$ so $$ \lim_{h\to0}\frac{f(a+h)-f(a+3h)}{2h}= \lim_{h\to0}\bigl(-f'(a)+\tfrac{1}{2}\varphi(h)-\tfrac{3}{2}\varphi(3h)\bigr)=-f'(a) $$