# If f′(x_0) = 5, then lim_{ℎ→0} (f(x_0+ℎ)−f(x_0−ℎ))/ℎ ?

## If f′(x_0) = 5," then "lim ℎ→0 {f(x_0+ℎ)−f(x_0−ℎ)}/ ℎ ?

Jun 23, 2018

10

#### Explanation:

Since ${f}^{'} \left({x}_{0}\right) = 5$, we have

${\lim}_{h \to 0} \frac{f \left({x}_{0} + h\right) - f \left({x}_{0}\right)}{h} = 5$

but this also means that

${\lim}_{h \to 0} \frac{f \left({x}_{0} - h\right) - f \left({x}_{0}\right)}{- h} = 5$

or

${\lim}_{h \to 0} \frac{f \left({x}_{0}\right) - f \left({x}_{0} - h\right)}{h} = 5$

Now

${\lim}_{h \to 0} \frac{f \left({x}_{0} + h\right) - f \left({x}_{0} - h\right)}{h}$
$= {\lim}_{h \to 0} \frac{f \left({x}_{0} + h\right) - f \left({x}_{0}\right) + f \left({x}_{0}\right) - f \left({x}_{0} - h\right)}{h}$
$= {\lim}_{h \to 0} \left[\frac{f \left({x}_{0} + h\right) - f \left({x}_{0}\right)}{h} + \frac{f \left({x}_{0}\right) - f \left({x}_{0} - h\right)}{h}\right]$
$= {\lim}_{h \to 0} \frac{f \left({x}_{0} + h\right) - f \left({x}_{0}\right)}{h} + {\lim}_{h \to 0} \frac{f \left({x}_{0}\right) - f \left({x}_{0} - h\right)}{h}$
$= 5 + 5 = 10$

Jun 23, 2018

$10$.

#### Explanation:

Note that, ${\lim}_{h \to 0} \frac{f \left({x}_{0} + h\right) - f \left({x}_{0}\right)}{h} = f ' \left({x}_{0}\right) \ldots \ldots \ldots \ldots . \left(1\right)$.

Also, $f ' \left({x}_{0}\right) = {\lim}_{x \to {x}_{0}} \frac{f \left(x\right) - f \left({x}_{0}\right)}{x - {x}_{0}}$.

Let, $x = {x}_{0} - h , \text{ so that, as } x \to {x}_{0} , h \to 0.$

$\therefore f ' \left({x}_{0}\right) = {\lim}_{h \to 0} \frac{f \left({x}_{0} - h\right) - f \left({x}_{0}\right)}{- h} ,$

$\mathmr{and} , {\lim}_{h \to 0} \frac{f \left({x}_{0} - h\right) - f \left({x}_{0}\right)}{h} = - f ' \left({x}_{0}\right) \ldots \ldots \ldots \ldots \left(2\right)$.

Subtracting $\left(2\right)$ from $\left(1\right)$, we find,

${\lim}_{h \to 0} \frac{\left\{f \left({x}_{0} + h\right) - f \left({x}_{0}\right)\right\} - \left\{f \left({x}_{0} - h\right) - f \left({x}_{0}\right)\right\}}{h} = 2 f ' \left({x}_{0}\right) ,$

$i . e . , {\lim}_{h \to 0} \frac{f \left({x}_{0} + h\right) - f \left({x}_{0} - h\right)}{h} = 2 f ' \left({x}_{0}\right)$.

$\Rightarrow \text{The Reqd. Lim.} = 2 \times 5 = 10$.