Question

Solve the following (limit; L'Hospital's Rule)?

`\lim_(x\rightarrow0^+)(\tan(2x))^x`

What I've tried
`y=(\tan(2x))^x`
`\rArr\ln(y)=\lim_(x\rightarrow0^+)\ln((\tan(2x))^x)...`
got stuck afterwards.

Answer 1

`1`

Explanation:

`(tan(2x))^x=(sin(2x))^x/(cos(2x))^x`

`sin(2x)= 2x-(2x)^3/(3!)+(2x)^5/(5!)+ cdots`

is an alternate series and for `-pi/2 le x le pi/2` verifies

`2x-(2x)^3/(3!) le sin(2x) le 2x`

then

`lim_(x->0)(2x-(2x)^3/(3!))^x le lim_(x->0)(sin(2x))^x le lim_(x->0)( 2x)^x`

now using the binomial expansion

`x^x = (1+(x-1))^x = 1 + x(x-1)+x(x-1)(x-1)^2/(2!)+x(x-1)(x-2)(x-1)^3/(3!)+cdots = 1+xp(x)` where `p(x)` represents a suitable polynomial. So

`lim_(x->0)x^x=lim_(x->0) 1+xp(x)=1`

Of course

`lim_(x->0)(2x)^x=(lim_(y->0)y^y)^(1/2)= 1^(1/2)=1`

Finally

`lim_(x->0)(tan(2x))^x=(lim_(x->0)(sin(2x))^x)/(lim_(x->0)(cos(2x))^x)=1/1^0= 1`

Answer 2

1

Explanation:

Let y = (tan 2x)^x, then

`lim x to 0_+- ln y`

`= lim x to 0_+- x ln tan 2x`

`= lim x to 0_+- (ln tan 2x)/(1/x)`

`=lim x to 0_+- (( ln tan 2x )')/((1/x)')`

`=lim x to 0_+- -2x^2sec^2 (2x)/tan (2x)`

`=lim x to 0_+-((2x)/(sin 2x))((x))/((cos2x)`

`=(1)(0)/1=0`. so,

`y to e^0 = 1`.