Lt (tanx/x)^(1/ x^2) as x nears zero?

You come till the last step as above,i.e,

Let L=Lim (tanx/x)^(1/x^2)

Than,

Ln L=Lim (1/x^2)Ln(tanx/x) (Ln for natural log, i.e.log to base e)

Now, as x->0, the numerator, Ln(tanx/x) goes to zero faster than the denominator, x^2(This is arrived at by showing that by expanding tan into ratio of sine and cosine series, tanx/x goes to one ans x goes to zero, linearly, i.e. at the same pace as x). Hence the limit goes to zero(This is an indirect use of L'Hospital's rule, by saying goes to zero faster, I mean differential is greater).

Thus,

Ln L=0 => L=1(answer)

see the problem in this question is that here the power is in the form of variable(which includes 'x'), and the procedure to solve these types of question is as follows

1. First let the question equals to 'y' (or any other variable as u wish).

2. "TAKE THE LOGaritm OF BOTH SIDES".

3. after that u know the property log a^x =x log a.

after folowilg this ur quetion will convert to

log y = lim 1/x^2[log(tanx/x)]

now as the limit x ---> 0 according to the question then the L hospital's rule is applicable on that

and solve the further question intil the difrential comes a determined value...

At the desk of: Abishek verma...

set x^3 -x² - 2x to 0 you may factor an x out x(x² - x - 2) factor out what's in parenthesis x(x - 2)(x + a million) set each and every x to 0 x=0 x - 2 = 0 so x = 2 x + a million = 0 so x = -a million

ƒ=Lt (tanx/x)^(1/ x²)[say]

Then, logƒ=Lt (1/x²)log(tanx/x)

=>logƒ=Lt [log(tanx)-log(x)]/x²

=>Use L.Hopital Rule Next.

Taking log is not a pre-defined property of the limit, but it has to be taken in this case.

Its one of the deepest mysteries of maths.

work work till you succeed.