Question

How to find the limit 5(1-cos2x)/sin2x as x approach 0 from the left ?

Answer 1

`lim_(x->0^-) (5(1-cos2x))/(sin2x) = 0`

Explanation:

To evaluate:

`lim_(x->0^-) (5(1-cos2x))/(sin2x)`

we can divide numerator and denominator by `2x` and then substitute `t= 2x`:

`lim_(x->0^-) (5(1-cos2x))/(sin2x) = 5 lim_(x->0^-) ((1-cos2x)/(2x))/((sin2x)/(2x)) = 5 lim_(t->0^-) ((1-cost)/t)/((sint)/t)`

Now consider the well known limits:

`lim_(x->0) sinx/x = 1`

`lim_(x->0) (1-cosx)/x = 0`

and we have:

`lim_(x->0^-) (5(1-cos2x))/(sin2x) = (5*0)/1 = 0`

graph{(5(1-cos(2x)))/sin(2x) [-10, 10, -5, 5]}

Answer 2

`0.`

Explanation:

We have, `1-cos2x=2sin^2x, and, sin2x=2sinxcosx.`

`:." The Reqd. Limit="lim_(x to 0){5(2sin^2x)}/{2sinxcosx}`

`=5{lim_(x to 0) tanx}`

`=5(tan0)`

`=0.`

Since the Limit exists, we have,

`lim_(xto0-){5(1-cos2x)}/(sin2x)=0=lim_(xto0+){5(1-cos2x)}/(sin2x)," too."`

Enjoy Maths.!