Question

Question #b5c34

Answer 1

We seek:

`lim_(x rarr 0) (tan2x-sin2x)/x^2 = 0`

Explanation:

We seek:

`L = lim_(x rarr 0) (tan2x-sin2x)/x^2`

If we put `x=09` we note both numerator and denominator are both `0`, so we have an indeterminate form `0/0`, and as such we can apply L'Hôpital's Rule to give:

`L = lim_(x rarr 0) (d/dx(tan2x-sin2x))/(d/dx x^2)`
`\ \ \ = lim_(x rarr 0) (2sec^2x-2cos2x)/(2x)`
`\ \ \ = lim_(x rarr 0) (sec^2x-cos2x)/(x)`

Again, we have an indeterminate form `0/0`, and as such we can apply L'Hôpital's Rule again to give:

`L = lim_(x rarr 0) (d/dx(sec^2x-cos2x))/(d/dx x)`
`\ \ \ = lim_(x rarr 0) (2secxsecxtanx+2sinx)/(1)`
`\ \ \ = 2lim_(x rarr 0) 2secx^2tanx+sinx`
`\ \ \ = 2(0+0)`
`\ \ \ = 0`

Answer 2

`0.`

Explanation:

Let us find the Limit L without using L'Hospital's Rule.

We will use `lim_(theta to 0) tantheta/theta=1.`

Knowing that, `tan 2x=(2tanx)/(1-tan^2x), &, sin2x=(2tanx)/(1+tan^2x),`

we have,

`tan2x-sin2x,`

`=(2tanx)/(1-tan^2x)-(2tanx)/(1+tan^2x),`

`=2tanx{{(1+tan^2x)-(1-tan^2x)}/((1-tan^2x)(1+tan^2x)}},`

`=(4tan^3x)/(1-tan^4x)xxx^3/x^3,`

`=4*(tanx/x)^3*(x^2*x)/(1-tan^4x).`

`rArr (tan2x-sin2x)/x^2=4*(tanx/x)^3*x/(1-tan^4x).`

`:. L=lim_(x to 0)4*(tanx/x)^3*x/(1-tan^4x),`

`=4*1^3*0/(1-0^4),`

`rArr L=0.`

Enjoy Maths.!