Expert: Paul Klarreich - 11/16/2005

Question
use l'hopital's rule to find the limits.

1) lim x-->0 (sin x)^(tan x)
2) lim x-->infinity (1+2x)^(1/2 ln x)
3) lim x-->0 sin 7x/tan 11x

Answer
Hi, Amy,

You're really moving fast.  Last I heard from you, it was basic calculus, like chain rule, etc.  Now you're up to indeterminate forms.  (That's what happens when you take a few days off.)

When you use l'Hospital's rule (by the way, when you write it, you have to either put that funny French mark over the 'o' or follow it by an 's', and don't forget to  capitalize the H.) you need an indeterminate form, like 0/0, of inf/inf, etc, and you will want to write a fraction.

1) For your first one, which approaches 0^0, also indeterminate, any time you have complicated exponentials, try looking for the limit of the logarithm of the function instead, then working back.

ln (sin x)^(tan x) = tan x ln(sin x) = ln(sin x)/cot x

Now differentiate top and bottom:
         cos x/sin x           cos           - 1
         ------------ = ----------------- = ----- =
         - cot x csc x  - sin cos csc/sin    csc x

= - sin x

Now as x --> 0, this approaches 0.
So if ln(our original function) --> 0, our original function --> 1.   (Remember, ln 1 = 0)

2) lim x-->inf (1+2x)^(1/2 ln x)

Now this obviously approaches infinity, since both your base (1+2x) and exponent are getting larger.  SOOOO.... I will assume you meant:

lim x-->infinity (1+2x)^(1/(2 ln x))

and just forgot to parenthesize.

We'll try the 'ln' stuff again:

ln( (1 + 2x)^(1/2 ln x)) = 1/(2 ln x) ln (1 + 2x) = ln(1 + 2x)/(2 ln x)
Now THAT approaches inf/inf, a nice indeterminate form, and we are in business (I hope)

Differentiate top and bottom:
2/(1 + 2x)       x        1
----------- = ------ = -------, which approaches 1/2.
  2/x        1 + 2x   1/x + 2

So ln(our original) --> 1/2, and therefore our original approaches exp(1/2), which is sqrt(e).

3) lim x-->0 sin 7x/tan 11x

This approaches 0/0, (good solid indeterminate form.)
D(sin 7x)     7 cos 7x
---------- =  ------------- = 7 cos 7x cos^2(11x)/11
D(tan 11x)    11 sec^2(11x)

As x -> 0, cos (ax) --> cos 0 = 1, so this is just 7/11