Show that limit of (4^x-4)/(x-1)=8log2 as x approaches to 1?

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Answer 1 · 2017-12-09T10:41:32

See below.

Using L'Hospital's Rule:

#d/dx(4^x-4)=4^xlog(4)=4^x*2log(2)#

#d/dx(x-1)=1#

#(4^x*2log(2))/1#

Plugging in 1:

#(4^(1)*2log(2))/1=8log(2)#

#lim_(x->1)(4^x-4)/(x-1)=8log(2)#