Question #d12dc

Question

1466 views around the world

You can reuse this answer
Creative Commons License

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.!