limit x to 0 sin^2 6x/4x tanx

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thilina-g | College Teacher | (Level 1) Educator

Posted on

`lim_(xto0) (sin^2(6x))/(4x xx tan(x))`

We can rewrite this as,

`lim_(xto0) sin(6x)/tan(x) xx lim_(xto0) sin(6x)/(4x)`

We can evaluate the two limits separately.

`lim_(xto0)sin(6x)/(4x) = 6/4 lim_(xto0)sin(6x)/(6x)`

If `x to 0` then, `6x to 0.`

Then,

`lim_(xto0)sin(6x)/(4x) = 6/4 lim_(6xto0)sin(6x)/(6x)`

`lim_(xto0)sin(6x)/(4x) = 6/4 lim_(6xto0)sin(6x)/(6x) = 6/4 xx 1`

Therefore,

`lim_(xto0)sin(6x)/(4x) = 6/4`

 

`lim_(xto0) sin(6x)/tan(x) = lim_(xto0) (2sin(3x)cos(3x))/tan(x)`

We can write `sin(3x)` as,

`sin(3x) = sin(2x+x)`

`sin(3x) = sin(2x)cos(x)+cos(2x)sin(x)`

`sin(3x) = 2sin(x)cos(x)cos(x) + cos(2x)sin(x)`

Therefore,

`lim_(xto0) sin(6x)/tan(x) = (2 xx (2sin(x)cos(x)cos(x) + cos(2x)sin(x)) xx cos(3x))/(sin(x)/cos(x))`

This will give,

`lim_(xto0) sin(6x)/tan(x) = (2 xx cos(x) xx(2sin(x)cos(x)cos(x) + cos(2x)sin(x)) xx cos(3x))/sin(x)`

`lim_(xto0) sin(6x)/tan(x) = 2 xx cos(x) xx(2cos(x)cos(x) + cos(2x)) xx cos(3x)`

 Now we can evaluate the limit.

`lim_(xto0) sin(6x)/tan(x) =2xx1xx(2xx1xx1+1)xx1`

` `

`lim_(xto0) sin(6x)/tan(x) =2 xx 3 xx 1 = 6`

 

Therefore,

 

`lim_(xto0) sin(6x)/tan(x) xx lim_(xto0) sin(6x)/(4x) = 6 xx 6/4 = 9`

 

Therefore,

`lim_(xto0) (sin^2(6x))/(4x xx tan(x)) = 9`

 

 

mlehuzzah's profile pic

mlehuzzah | Student, Graduate | (Level 1) Associate Educator

Posted on

Slightly modified from the previous poster:

Separate the limit into a product of two limits, as the previous poster did.

The limit as x->0 of (sin 6x)/4x = 6/4, argued as the previous poster did.

But to evaluate the second limit, (sin 6x)/(tan x), it would perhaps be easier to use l'Hopital's rule:

lim (sin 6x)/(tan x) = lim (6 cos 6x)/(sec^2 x) = (6 cos 0)/(sec^2 0) = 6

(For that matter, you can use l'Hopital's rule from the very beginning, without splitting it into a product, but the derivatives get quite messy along the way)

 

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