Question

Evaluate the limit `lim_(h rarr 0) (1-cos h)/h^2`?

Answer 1

We seek:

`L = lim_(h rarr 0) (1-cos h)/h^2`

Method 1 - L'Hôpital's rule

Both the numerator and the denominator `rarr 0` as `x rarr 0`. thus the limit `L` (if it exists) is of an indeterminate form `0/0`, and consequently, we can apply L'Hôpital's rule to get:

`L = lim_(h rarr 0) (d/(dh)(1-cos h))/(d/(dh) h^2) = lim_(h rarr 0) (sinh)/(2h)`

Again, we have an indeterminate form, so we can apply L'Hôpital's rule again:

`L = lim_(h rarr 0) (d/(dh) sin h)/(d/(dh) 2h) = lim_(h rarr 0) (cos h)/(2)`

And we can now directly evaluate this limit:

`L = (cos0)/2 = 1/2`

Method 2 - Trigonometric Manipulation

`L = lim_(h rarr 0) (1-cos h)/h^2`
`\ \ = lim_(h rarr 0) (1-cos h)/h^2 * (1+cos h)/(1+cos h)`
`\ \ = lim_(h rarr 0) ( (1-cos h)(1+cos h) ) /( h^2(1+cos h) )`
`\ \ = lim_(h rarr 0) ( (1-cos^2h) ) /( h^2(1+cos h) )`
`\ \ = lim_(h rarr 0) ( sin^2h ) /( h^2(1+cos h) )`
`\ \ = lim_(h rarr 0) ( sin^2h )/h^2 * 1/(1+cos h )`
`\ \ = lim_(h rarr 0) ( sin h )/h * ( sinh )/h * 1/(1+cos h )`
`\ \ = {lim_(h rarr 0) ( sin h )/h} * {lim_(h rarr 0) ( sinh )/h }* {lim_(h rarr 0) 1/(1+cos h )}`
`\ \ = {lim_(h rarr 0) ( sin h )/h}^2 * {lim_(h rarr 0) 1/(1+cos h )}`

And we note the first limit is a standard Calculus Trigonometry limit:

`lim_(x rarr 0) (sinx)/x = 1`

And the second limit can be calculated directly, leading to the desired result:

`L = {1}^2 * {1/(1+cos 0 )}`
`\ \ = 1 * 1/(1+1)`
`\ \ = 1/2`