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 ?#

2 Answers
Ananda Dasgupta
Jun 23, 2018

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#

Ratnaker Mehta
Jun 23, 2018

# 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#.

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