A few days ago

Pr0ject

Finding limit lim h->0 (4^h-1)/h?

says to find it in your calc and the answer is ln 4

does anyone know how!

Top 2 Answers

A few days ago

Anonymous

Favorite Answer

the limit comes from finding the derivative of 4^x from first principles

substitute a number near 0

like 0.00001

gives 1.386303

and e^1.386303 = 4.00003456

the smaller the number you try, the closer to

ln4 ≈ 1.386294361….

it gets

so you can then guess that

lim h->0 (4^h-1)/h = ln4

A few days ago

Anonymous

Substitute:

x = 4^h-1

Then

h = ln(1+x)/ln4 and x –> 0 as h –> 0

We reduced the initial limit to:

lim x–>0 xln4/ln(1+x)= ln4 lim x/ln(1+x)

= ln4 lim 1/ln[(1+x)^(1/x)]

= ln 4

Here the well known limit:

lim x–>0 (1+x)^(1/x) = e

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