Does the limit #lim_(x-3) (f(x)-f(3))/(x-3)# always exist?

Question

Does the limit #lim_(x->3) (f(x)-f(3))/(x-3)# always exist?

Charles Anctil · Accepted Answer

Recall the limit definition of the derivativbe, that is:

# f'(a) = lim_(x rarr a) (f(x)-f(a))/(x-a)#

We have:

# L = lim_(x->3) (f(x)-f(3))/(x-3)#

And so clearly:

# L = f'(3) #

Without further knowledge of the function we cannot determine if the limits exist. If it were known that #f(x)# was differentiable over some domain that included #x=3# then we could conclude that the limit exists.

Ezra Adams · Answer

Kindly refer to the Discussion given in the Explanation.

The Limit under reference may or may not exist. Its existence depends upon the definition of the function #f.# Consider the following Examples : # (1) : f(x)=x, x in RR.# Clearly, the Limit =#lim_(x to 3) {f(x)-f(3)}/(x-3),# #=lim_(x to 3)(x-3)/(x-3)............[because, f(3)=3]# #=lim_(x to 3) 1.# We find that, #lim_(x to 3) {f(x)-f(3)}/(x-3),# exists, and, is #1.# # (2) : f(x)=|x-3|, x in RR.# Remember that, # AA x in RR, |x|=x; if x>=0, &, |x|=-x, if x < 0.# Since, #f(3)=|3-3|=0,# we have, #{f(x)-f(3)}/(x-3)=|x-3|/(x-3).# #"Now, as "x to 3-, x < 3 :. (x-3) <0.# #:. |x-3|=-(x-3).# #:. lim_(x to 3-){f(x)-f(3)}/(x-3),# #=lim_(x to 3-) {-(x-3)}/(x-3),# #rArr lim_(x to 3-) {f(x)-f(3)}/(x-3)=-1....................(star^1).# On the other hand, as #x to 3+, x>3. :. |x-3|=(x-3).# # rArr lim_(x to 3+){f(x)-f(3)}/(x-3)=1..........................(star^2).# #(star^1), &, (star^2),# #rArr lim_(x to 3-) {f(x)-f(3)}/(x-3)=-1!=1=lim_(x to 3+){f(x)-f(3)}/(x-3).# We conclude that, #lim_(x to 3){f(x)-f(3)}/(x-3)# does not exist. Enjoy Maths.!

William Abdalla · Answer

undefined The limit $ \lim_{x \to 3} \frac{f(x) - f(3)}{x - 3} $ will exist if the function $ f(x) $ is continuous at $ x = 3 $. If $ f(x) $ is continuous at $ x = 3 $, then the limit will exist. If $ f(x) $ is not continuous at $ x = 3 $, then the limit may or may not exist, depending on the behavior of $ f(x) $ near $ x = 3 $.

Does the limit #lim_(x-&gt;3) (f(x)-f(3))/(x-3)# always exist?

Does the limit #lim_(x->3) (f(x)-f(3))/(x-3)# always exist?