Question

Evaluate? a) lim(x tends to 0)sin5x/tan3x

Answer 1

`L=5/3`

Explanation:

We know that

`color(red)((1)lim_(x to0)sintheta/theta=1 and lim_(x to0)cosx=1`

Here,

`L=lim_(x to0)(sin5x)/(tan3x)`

`=lim_(x to0)(sin5x)/((sin3x)/(cos3x))`

`=lim_(x to0)(sin5x)/(sin3x)xxlim_(x to0)(cos3x)`

`=lim_(x to0)[(sin(5x)/(5x))/(sin(3x)/(3x))]xx5/3xx(cos0)..toApply (1)`

`=(lim_(5x to0)sin(5x)/((5x)))/(lim_(x to0)sin(3x)/((3x)))xx5/3xx1`

`=(1)/(1)xx5/3`

`L=5/3`

Answer 2

`lim_(x->0)(sin5x)/(tan3x)=5/3`

Explanation:

`lim_(x->0)(sin5x)/(tan3x)`

= `lim_(x->0)(sin5x)/(5x) * cos3x * (3x)/(sin3x)*(5x)/(3x)`

= `lim_(x->0)((sin5x)/(5x))*lim_(x->0)(cos3x)*lim_(x->0)((3x)/(sin3x))*lim_(x->0)((5x)/(3x))`

Now as `x->0`, `5x->0` and `3x->0` and hence, we can write this as

`lim_(5x->0)((sin5x)/(5x))*lim_(3x->0)(cos3x)*lim_(3x->0)((3x)/(sin3x))*5/3`

= `1*1*1*5/3`

= `5/3`