evaluate the limit?

Question

lim (( ln (4+7x) - ln 4 ) / (e^4x -1)) ...ffrom the answer key: its 7/16
x->0
how do u do this in the case of a ln?
im confused please help

Answer 1

since Lim = 0/0 if you plug 0 in for x, use L'Hospital rule.

Lim f(x) / h(x) = d/dx f(x) / d/dx g(x)
x > c

Lim = d/dx [ ln(4 + 7x) - ln(4)] / d/dx [ e^(4x) - 1]
x > 0

use the chain rule:
let u = 4 + 7x
then d/du ln(u) = 1/u
d/dx 4 + 7x = 7

use the chain rule again:
let v = 4x
then d/dv (e^v) = e^v
d/dx (4x) = 4

Lim = [1/(4 + 7x) * 7] / [e^(4x) * 4]
x > 0

Lim = [7/(4+7x)] / [4e^(4x)]
Lim > 0

plug 0 in
Lim = [ 7 / (4 + 7*0)] / [4e^(4*0)]
x > 0

Lim = (7/4) / [4e^(0)]
x > 0

Lim = (7/4) / (4*1)
x > 0

lim = (7/4) / 4
x > 0

Lim = (7/4) * (1/4)
x > 0

Lim = 7/16
x > 0


hope it helps
Rec

Answer 2

lim A / B = lim A' / B'

lim (( ln (4+7x) - ln 4 ) / (e^4x -1)) =
x->0

lets A = ln (4+7x) - ln 4 --> A' = 7 / (7x + 4)
lets B = e^4x -1 --> B' = 4 e^4x

so, the limit will be

lim (7 / (7x + 4)) / 4e^4x
x->0

insert x = 0

= (7/4) / 4 = 7/16

solved

Answer 3

When you put x = 0 into the expression you get 0/0, so this is a candidate for L'Hospital's rule. Take the derivatives of the numerator and denominator:

Numerator: [1/(4+7x)]*7
Denominator: 4*e^4x

Now put in x = 0

Numerator: 7/4
Denomintaor: 4

so the answer is (7/4) / 4 or 7/16