How do you prove that #h(x) = sqrt(2x - 3)# is continuous as x=2?

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Answer 1 · 2018-04-12T01:01:01

A function #f(x)# is said to be continuous at #a# if #lim_(xrarra)f(x)=f(a)#

So in this case, our #f(x)# is #sqrt(2x-3)# and #a# is #2#. Compute the limit, then verify that it equals #h(2)#:

#color(white)=lim_(xrarr2)sqrt(2x-3)#

#=sqrt(2(2)-3)#

#=sqrt(4-3)#

#=sqrt1#

#=1#

Here's #h(2)#:

#h(2)=sqrt(2(2)-3)#

#color(white)(h(2))=sqrt(4-3)#

#color(white)(h(2))=sqrt1#

#color(white)(h(2))=1#

Since the limit and the function are equal, the function #h(x)# is continuous at #x=2#. Hope this helped!