Bonus Question Attempt from Calculus Final

[RadicalAlchmy]

This was the final bonus question on my first university and I made a serious attempt at it. I know this isn't technically a homework question, so I will understand if it goes unanswered, by founding out if I did this correctly would go a long way to alleviating some of my final exam anxieties.

Homework Statement

f(x)= (1/x)∫₀^x((1-tan(2t))^1/t)dt when x≠0 and k when x=0

Find the value of k that makes f(x) continuous

Homework Equations

L'Hopital's Rule, definition of continuity

The Attempt at a Solution

lim_{x -> 0} (1/x)∫₀^x((1-tan(2t))^1/t)dt = k

lim_{x -> 0} (∫₀^x((1-tan(2t))^1/t)dt)/x = k

By direct substitution, lim_{x -> 0} (∫₀^x((1-tan(2t))^1/t)dt)/x = 0/0

Since this is an indeterminant form, apply L'Hopital's Rule.

lim_{x -> 0} ((1-tan(2x))^1/x)/1 = k

lim_{x -> 0} ln( ((1-tan(2x))^1/x) ) = ln k

lim_{x -> 0} (1/x)ln( 1-tan(2x) ) = ln k

lim_{x -> 0} ln( 1-tan(2x) )/x = ln k

By direct substition again, lim_{x -> 0} ln( 1-tan(2x) )/x = 0/0.
Apply L'Hopital's Rule.

lim_{x -> 0} -(2)sec²(2x)/(1-tan(2x) = ln k

Direct substition:

-(2)sec²(0)/(1-tan(0)) = ln k

-2 = ln k

k = e^-2

Thanks in advance.

 

[lurflurf]

That looks right, the answer is right. The short version is
The integral does not matter (it is the average near zero so just the limit)
lim (1-tan(2t))^(1/t)=e^ lim -tan(2t)/t
this follows from lim (1-x)^(1/x)=1/e