Question

How do you find the limit of `f(x)=(secx-1)/x^2` as x approaches 0?

Answer 1

The answer is: `1/2`.

`lim_(xrarr0)(secx-1)/x^2=lim_(xrarr0)(1/cosx-1)/x^2=`

`lim_(xrarr0)((1-cosx)/cosx)/x^2=lim_(xrarr0)(1-cosx)/(cosx*x^2)=`

`=lim_(xrarr0)1/cosx*(1-cosx)/x^2=1/1*1/2=1/2`,

since

`lim_(xrarr0)(1-cosx)/x^2=1/2` is a particular memorable limit.

Answer 2

`1/2`

Explanation:

`lim_(x to 0)(secx-1)/x^2=lim_(x to0)(1/cosx-1)/x^2=`

`lim_(x to 0)((1-cosx)/cosx)/x^2=lim_(x to0)(1-cosx)/(cosx*x^2)=`

`=lim_(x to 0)1/cosx*(1-cosx)/x^2`

`=1/cosx* lim_(x to 0)(1-cosx)/x^2`

as `cos x = 1` and is continuous at `x = 0`

`=lim_(x to 0)(1-cosx)/x^2`

now `cos x = 1 - 2 sin^2( x/2)`

`= lim_(x to 0)(1-( 1 - 2 sin^2 (x/2)))/x^2`

`=lim_(x to 0)(2 sin^2 (x/2))/x^2`

`= 2lim_(x to 0)( sin (x/2))/x * ( sin (x/2))/x`

`= 1/2 lim_(x to 0)( sin (x/2))/(x/2) * ( sin (x/2))/(x/2)`

with `u = x/2`

`= 1/2 lim_(u to 0) ( sin (u))/(u) * ( sin (u))/(u)`

`= 1/2 * 1 * 1`

relying on fundamental limit `lim_{x to 0} (sin u}/u = 1`