How do you find the limit of #(2^w-2)/(w-1)# as w approaches 1?

Question

1728 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-21T22:44:27

#ln4#

We want to evaluate this: #lim_(wrarr1)(2^w-2)/(w-1)#

Putting #w=1# directly gives us: #(2^1-2)/(1-1)=0/0# Which is an indeterminate form. So, we a free to apply Hospital's rule

Hospital's rule simply says that we differentiate top and bottom of the #lim# until the indeterminate form disappears.

#lim_(wrarr1)(2^w-2)/(w-1) " "# becomes #" "lim_(wrarr1)(2^wln2)/(1)=2^1ln2=2ln2=ln2^2=color(blue)ln4#

#color(white)#

If you are curious as to how I differentiated #color(red)(2^w)# then you may read on!

Suppose, #y=2^w " "# then #" "lny=wln2#

Differentiating implicitly we get, #" "1/y(dy)/(dx)=ln2#

#=>(dy)/(dx)=yln2=2^wln2#