Question

Question #5f7a4

Answer 1

This limit can be found using l'Hopital's rule or by using knowledge of trigonometry.

Explanation:

`lim_(xrarr0) (1-cosx)/x^2` has indeterminate form `0/0`

If you try rewriting to get a fundamental trigonometric limit, you'll get:

`lim_(xrarr0) ((1-cosx)/x * 1/x)`

This limit has indeterminate form: `0 * oo`.

If you have learned it, you can use l'Hopital's rule on the original ratio form.

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

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Without l'Hopital

Use a 'trick' you may recall from trigonometry. (It shows up in some identity problems.)

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

`= (1-cos^2x)/(x^2(1+cosx))`

`= sin^2x/(x^2(1+cosx))`

Now, we can split this into factors whose limits we can find.

`= sinx/x sinx/x 1/(1+cosx)`

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

`= (1)(1)(1/(1+1))`

`= 1/2`