Question

What is equal ? `lim_(x->pi/2)sin(cosx)/(cos^2(x/2) -sin^2(x/2) )= ?`

Answer 1

`1`

Explanation:

`"Note that : "color(red)(cos^2(x)-sin^2(x) = cos(2x))`
`"So here we have"`

`lim_{x->pi/2} sin(cos(x))/cos(x)`

`"Now apply rule de l' Hôptial :"`

`= lim_{x->pi/2} cos(cos(x))*(-sin(x))/(-sin(x))`
`= lim_{x->pi/2} cos(cos(x))`
`= cos(cos(pi/2))`
`= cos(0)`
`= 1`

Answer 2

`1`.

Explanation:

Here is a way to find the limit without using L'Hospital's Rule :

We will use, `lim_(alpha to 0)sinalpha/alpha=1`.

If we take `cosx=theta`, then as `x to pi/2, theta to 0`.

Replacing `cos^2(x/2)-sin^2(x/2)` by `cosx=theta,` we have,

`:."The reqd. lim."=lim_(theta to 0)sintheta/theta=1`.

Answer 3

`1`

Explanation:

We know that,
`color(red)(cosA=cos^2(A/2)-sin^2(A/2))`
So,
`L=lim_(x->pi/2)(sin(cosx))/(cos^2(x/2)-sin^2(x/2))=lim_(x->pi/2)(sin(cosx))/(cosx)`
Take,`cosx=theta,`
We get, `xto(pi/2)rArrtheta tocos(pi/2)rArrtheta to0.`
`:.L=lim_(theta->0)(sintheta)/theta=1`