limit question?

Question

How do I solve this using l'hopital

lim e^{ln[tan(x)/(x)]/[(x)^2]}
x->0

it reads:

e is the base

this is the power: ln[tan(x)/(x)]/[(x)^2]
ln[tan(x)/(x)] is the numerator
[(x)^2] is the denominator


if you know how to solve this please show all your work!
THANKS!

Answer 1

[x→0]lim e^(ln (tan x/x) / x²)

First, since e^x is continuous, we can move the limit inside the exponential:

e^([x→0]lim ln (tan x/x) / x²)

Next, note that since [x→0]lim tan x/x = [x→0]lim sec² x = 1, we have [x→0]lim ln (tan x/x) = 0, so this is a 0/0 form. Invoke L'hopital:

e^([x→0]lim d(ln (tan x/x))/dx / (2x))

Using the chain and quotient rules:

e^([x→0]lim 1/(tan x/x) d(tan x/x)/dx / (2x))
e^([x→0]lim 1/(tan x/x) (x sec² x - tan x)/x² / (2x))

Simplifying:

e^([x→0]lim x/tan x (x sec² x - tan x)/(2x³))

The limit of a product is the product of the limits:

e^(([x→0]lim x/tan x) ([x→0]lim (x sec² x - tan x)/(2x³)))

And we already know [x→0]lim x/tan x = 1, so this is:

e^([x→0]lim (x sec² x - tan x)/(2x³))

That numerator is ugly. Extract a factor of sec² x:

e^([x→0]lim sec² x (x - sin x cos x)/(2x³))

Again, the limit of a product is the product of the limits:

e^(([x→0]lim sec² x) ([x→0]lim (x - sin x cos x)/(2x³)))
e^([x→0]lim (x - sin x cos x)/(2x³))

Use a trig identity to simplify further:

e^([x→0]lim (x - sin (2x)/2)/(2x³))

This limit is a 0/0 form. Invoke L'hopital again:

e^([x→0]lim (1 - cos (2x))/(6x²))

Still 0/0. Invoke L'hopital again:

e^([x→0]lim 2 sin (2x)/(12x))

And once more:

e^([x→0]lim 4 cos (2x)/12)

And finally we have something we can evaluate:

e^(4/12)
∛e

And we are done.

Answer 2

Well, you can cancel out the natural log ("e" and "ln"). To do this, you will have to split up the "ln[tan(x)/(x)]/[(x)^2]" part.

It can then be written like this:
ln[tan(x)/(x)] - ln[(x)^2] since they are divided. I forgot the exact rule that makes this possible...but can google Natural Log Math rules and should get some hits.

This then makes this equation from above:
lim e^{ln[tan(x)/(x)] - ln[(x)^2]}

which can be further split into two more limits:
lim e^{ln[tan(x)/(x)]} - lim e^{ ln[(x)^2]}

Now you can cancel out the natural logs and use l'hopital rule:
lim tan(x)/(x) - lim x^2

lim tan(x)/(x) = sec^2(x)
lim x^2 = 2x

Therefore lim e^{ln[tan(x)/(x)]/[(x)^2]} = sec^2(x) - 2x

Since the limit aproaches "0" the answer should be 1 since sec^2(0) = 1 and 2(0) = 0.

Hope that helps!

Answer 3

A quicker way to do the problem is to use Taylor expansion in which,
tan(x) = x + x^3/3 +...
ln(1+x) = x - x^2/2 + ...
As x->0, we have
tan(x)/(x) ~ 1 + x^2/3
ln(tan(x)/(x))/x^2 ~ ln(1 + x^2/3)/x^2 ~ 1/3
Therefore,
lim e^{ln[tan(x)/(x)]/[(x)^2]} = e^(1/3)
x->0
---------
Ideas to save time: Use Taylor expansion. Check my comments in your previous question, if you like this approach.

Answer 4

e^{ln[tan(x)/(x)]/[(x)^2]}
= e^ ln [(tan x / x)]^[1/x^2]
= (tan x / x)^(1 / x^2)
= [ 1 + (tan x - x) / x ] ^ [ [ x / (tanx - x) ] * [ (tanx - x) / x^2 ] ]

lim (x -> 0) e^{ln[tan(x)/(x)]/[(x)^2]}
= lim (x -> 0) [ 1 + (tan x - x) / x ] ^ [ [ x / (tanx - x) ] * [ (tanx - x) / x^3 ] ]
= e ^ lim (x -> 0) [ (tanx - x) / x^3 ]
(because, lim (x -> 0) (tan x - x) / x = lim (x -> 0) tanx / x - 1 = 0
Now, using L'Hospital's theorem,
= e^ lim (x -> 0) [(sec^2 x - 1) / 3x^2]
= e^ lim (x -> 0) [ (2sec x tan x ) / 6x ]
= e^ (1/3)