Question #bc838

1 Answer
Nov 17, 2017

Answer:

We can't, because that is not the limit. If you intended to write #lim_(x->10) 3-(4x)/5 = -5#, then see explanation below.

Explanation:

One of the first steps in checking the limit of a function at an x-value is seeing if the function itself is defined for that value, followed by whether it will be defined for nearby x-values.

In this case our function is:

#(3-4x)/5#

Because this is a simple linear function, it is self evident that the function will be continuous throughout. Thus, the limit shall exist everywhere, and will be equal to the value #f(x_o)# for any #x_0# in the domain.

Knowing this, we plug in #x=10#

#(3-4(10))/5 = (3-40)/5 = -37/5 = -7 2/5 = -7.4#

This is not equal to the #-5# you sought, ergo, #-5# is not the appropriate limit.

It is possible that you mis-wrote the function, but for the function as you have written it, you cannot prove #-5# is the limit because it is not the limit.

If you instead meant to write:

#3-(4x/5)#

You can perform the same process as above. Again, since the function is linear and continuous throughout, the function will be equal to its own limit at any point.

#f(10) = 3 - (4(10))/5 = 3 -40/5 = 3-8 = -5#