Question

How do you find `\lim _ { x \rightarrow - \frac { x } { 4} } \frac { \cos 2x } { \sin x + \cos x }`?

Answer 1

`lim_(x->-pi/4)cos(2x)/(sin(x)+cos(x))=sqrt2`

Explanation:

First investigate: Substitute `-pi/4` for `x`

Since:

`color(blue)(sin(-pi/4)=-sqrt2/2`

`color(blue)(cos(-pi/4)=sqrt2/2`

`lim_(x->-pi/4)cos(2x)/(sin(x)+cos(x))=0/(-sqrt2/2+sqrt2/2)=0/0`

We end up with the indeterminate form `0/0` which implies we should use L'Hospitals Rule

This rule states:

If, `lim_(x->a)f(x)/g(x)=0/0`

Then, `lim_(x->a)f(x)/g(x)=lim_(x->a)(f'(x))/(g'(x))`

So,

`lim_(x->-pi/4)cos(2x)/(sin(x)+cos(x))=lim_(x->-pi/4)(d/dx[cos(2x)])/(d/dx[(sin(x)+cos(x)]`

`=lim_(x->-pi/4)(-2sin(2x))/(cos(x)-sin(x))`

Substitute in `-pi/4` for `x`:

`=lim_(x->-pi/4)(-2sin(2(color(red)(-pi/4))))/(cos(color(red)(-pi/4))-sin(color(red)(-pi/4)))=(-2*-1)/(sqrt2/2-(-sqrt2/2))`

`=2/((2sqrt2)/2)=2/sqrt2`

Simplify by applying the exponent rule: `x^a/x^b=x^(a-b)`

`2/sqrt2=2^(1-1/2)=2^(1/2)=sqrt2`

Answer 2

Use `cos2x=cos^2x-sin^2x`, then factor and reduce.

Explanation:

`lim_(xrarr-pi/4)(cos2x)/(sinx+cosx) = lim_(xrarr-pi/4)(cos^2x-sin^2x)/(cosx+sinx)`

`= lim_(xrarr-pi/4)((cosx-sinx)(cosx+sinx))/(cosx+sinx)`

`= lim_(xrarr-pi/4)(cosx-sinx))`

`= cos(-pi/4) - sin(-pi/4)`

`= sqrt2/2-(-sqrt2/2) = sqrt2`