Show that f(x)=2a+3b is continous, where a and b are constants?

1 Answer
F. Javier B.
Mar 29, 2018

See explanation below

Explanation:

A function is called continuous in #x_0# if

#f(x_0)# exists and is finite

#lim_(xtox_0)f(x)# exists and finally

#lim_(xtox_0)f(x)=f(x_0)#

In our case #f(x)=2a+3b# with #a# and #b# constants. So the function doesn't depend of x and for this reason is constant for every value of x, and his value is #f(x)=2a+3b# whatever be a and b.

For this reason

#f(x_0)=2a+3b# exists for every #x_0# and is finite
#lim_(xtox_0)f(x)=2a+3b#. This proof that #f(x)# is continous in all #x in RR#

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