Question #d12dc

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Answer 1 · 2017-11-14T06:25:08

# 8ln2, or, ln(2^8)=ln256.#

Let us solve this Problem without using L'Hospital's Rule.

We need this Standard Limit :

#lim_(h to 0) (a^h-1)/h=lna, a in RR^+ -{1}...............(star).#

#"The Reqd. Lim.="lim_(x to 3)(2^x-8)/(x-3).#

Subst. #x=3+h," so that, as "x to 3, h to 0.#

#"Therefore, the Reqd. Lim.="lim_(h to 0)(2^(h+3)-8)/h,#

#=lim(2^h*2^3-8)/h,#

#=lim(8*2^h-8)/h,#

#=lim{8((2^h-1)/h)},#

#=8ln2, or, ln(2^8)=ln256.............[because, (star).].#

Enjoy Maths.!