Question

Question #eefeb

Answer 1

By using trig identities and algebraic manipulation, together with L'Hospitals Rule and other limit laws, we eventually get this limit equal to `1/2`.

Explanation:

By a trig identity, `sin2x=2sinxcosx`

Taking out a common denominator and simplifying algebraically, we get

`lim_(x->0)[x/(2sinxcosx)-sinx] = lim_(x->0)((x-2sin^2xcosx)/(2sinxcosx))`

This is an indeterminant limit form of type `0/0` and therefore we may use L'Hospital's Rule to evaluate it to obtain

`lim_(x->0)((x-2sin^2xcosx)/(2sinxcosx))= lim_(x->0)[(d/dxx-2sin^2xcosx)/(d/dx2sinxcosx)]`

`=lim_(x->0)((1-sin^3x-4sinxcos^2x)/(2cos^2x-2sinxcosx))`

`=1/2`

Answer 2

See the explanation section.

Explanation:

To find `lim_(xrarr0)(x/sin(2x) - sinx)`,

we'll need the fundamental trigonometric limit: `lim_(thetararr0)sintheta/theta = 1`

in the reciprocal form:
`lim_(thetararr0)theta/sintheta = 1`

`lim_(xrarr0)(x/sin(2x) - sinx) = lim_(xrarr0)(x/sin(2x)) - lim_(xrarr0)sinx`

`= lim_(xrarr0)(x/sin(2x))-0`

So the challenge is finding `lim_(xrarr0)x/sin(2x)`

Solution 1
`lim_(xrarr0)x/sin(2x) = lim_(xrarr0)1/2 (2x)/sin(2x)`

`= 1/2 lim_(xrarr0) ((2x))/sin(2x)`

`= 1/2 (1) = 1/2`

We used the fact that as `xrarr0`, we also have `2xrarr0`. We could formalize the substitution `theta = 2x` is we want to be precise.

(The technique in solution 1 also works for `lim_(xrarr0)x/sin(5x) = 1/5`.)

Solution 2

Use `sin(2x) = 2sinxcosx` to rewrite.

`lim_(xrarr0)x/sin(2x) = lim_(xrarr0)x/(2sinxcosx)`

`= lim_(xrarr0)x/(2sinxcosx)`

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

`= 1/2[lim_(xrarr0)x/sinx][lim_(xrarr0)1/cosx]` (if both limits exist)

`= 1/2[1][1/1] = 1/2`

(Unless you know the formula for `sin(5x)`, this solution method will not work for `lim_(xrarr0)x/sin(5x)`.)