Question

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 ?`

Answer 1

10

Explanation:

Since `f^'(x_0) = 5`, we have

`lim_{h to 0} {f(x_0+h)-f(x_0)}/h = 5`

but this also means that

`lim_{h to 0} {f(x_0-h)-f(x_0)}/(-h) = 5`

or

`lim_{h to 0} {f(x_0)-f(x_0-h)}/h = 5`

Now

`lim_{h to 0} {f(x_0+h)-f(x_0-h)}/h`
`=lim_{h to 0} {f(x_0+h)-f(x_0)+f(x_0)-f(x_0-h)}/h`
`=lim_{h to 0} [{f(x_0+h)-f(x_0)}/h+ {f(x_0)-f(x_0-h)}/h]`
`=lim_{h to 0} {f(x_0+h)-f(x_0)}/h+lim_{h to 0} {f(x_0)-f(x_0-h)}/h`
`=5+5 = 10`

Answer 2

`10`.

Explanation:

Note that, `lim_(h to 0){f(x_0+h)-f(x_0)}/h=f'(x_0).............(1)`.

Also, `f'(x_0)=lim_(x to x_0){f(x)-f(x_0)}/(x-x_0)`.

Let, `x=x_0-h," so that, as "x to x_0, h to 0.`

`:. f'(x_0)=lim_(h to 0){f(x_0-h)-f(x_0)}/(-h),`

`or, lim_(h to 0){f(x_0-h)-f(x_0)}/h=-f'(x_0)............(2)`.

Subtracting `(2)` from `(1)`, we find,

`lim_(h to 0)[{f(x_0+h)-f(x_0)}-{f(x_0-h)-f(x_0)}]/h=2f'(x_0),`

`i.e., lim_(h to 0){f(x_0+h)-f(x_0-h)}/h=2f'(x_0)`.

`rArr"The Reqd. Lim."=2xx5=10`.