Determine whether given statements about a function differentiable at a point are true or false

by RoRi

September 19, 2015

Let $f$ be a function and $a$ a point such that $f'(a)$ exists. Determine which of the following are true or false.

$\displaystyle{f'(a) = \lim_{h \to a} \frac{f(h) - f(a)}{h-a}}$ .
$\displaystyle{f'(a) = \lim_{h \to 0} \frac{f(a) - f(a -h)}{h}}$ .
$\displaystyle{ f'(a) = \lim_{t \to 0} \frac{f(a+2t) - f(a)}{t}}$ .
$\displaystyle{ f'(a) = \lim_{t \to 0} \frac{f(a+2t) - f(a+t)}{2t}}$ .

True. Let $k = h-a$ . Then $k \to 0$ as $(h-a) \to 0$ implies $k \to 0$ as $h \to a$ . Thus,
$f'(a) = \lim_{k \to 0} \frac{f(a+k) - f(a)}{k} = \lim_{h \to a} \frac{f(a+h-a) - f(a)}{h-a} = \lim_{h \to a} \frac{f(h) - f(a)}{h-a}. \qquad \blacksquare$
True. Let $k = -h$ , then
$\lim_{h \to 0} \frac{f(a) - f(a-h)}{h} = \lim_{k \to 0} \frac{f(a) - f(a+k)}{-k} = \lim_{k \to 0} \frac{f(a+k) - f(a)}{k} = f'(a). \qquad \blacksquare$
False. First, from the definition of limit we can conclude
$\lim_{t \to 0} f(t) = \lim_{2t \to 0} f(t)$

(by taking $\delta' = \frac{\delta}{2}$ ). Then we have,

$\begin{align*} \lim_{t \to 0} \frac{f(a+2t) - f(a)}{t} &= \lim_{2t \to 0} \frac{f(a+2t) - f(a)}{t} \\ &= 2 \lim_{2t \to 0} \frac{f(a+2t) - f(a)}{2t} \\ &= 2 f'(a) \neq f'(a) & \text{if } f'(a) \neq 0. \end{align*}$

Thus, this is not true in general (since there are many function differentiable at a point $a$ such that the derivative at that point is nonzero). ( Note: This statement would be true were the denominator in the given equation $2t$ instead of $t$ .)
False. Here we evaluate,
$\begin{align*} \lim_{t \to 0} \frac{f(a+2t) - f(a+t)}{2t} &= \frac{1}{2} \left( \lim_{t \to 0} \frac{f(a+2t)-f(a)}{t} - \lim_{t \to 0} \frac{f(a+t) - f(a)}{t} \right) \\ &= \frac{1}{2} ( 2f'(a) - f'(a)) \qquad (\text{part (c)})\\ &= \frac{1}{2} f'(a) \neq f'(a) \qquad \text{if } f'(a) \neq 0. \end{align*}$

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One comment

December 31, 2022 at 3:57 pm

William says:

I think another way to do part d would be to let k = a+t.

We have 1/2 times the regular limit function (with k’s instead of a’s), and if we note that v -> a as t -> 0 we get 1/2 f'(a), which is not equal to f'(a)

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