# What is  `lim_(x-gtoo) (3sqrt(x^4-3x))/(2x^2+cosx)`

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

Posted on

`lim_(x-gtoo) (3sqrt(x^4-3x))/(2x^2+cosx)`

Let's divide both numerator and denominator by `x^2` .

= `lim_(x-gtoo) ((3sqrt(x^4-3x))/x^2)/((2x^2+cosx)/x^2)`

= `lim_(x-gtoo) (3sqrt(x^4/x^4-(3x)/x^4))/(2+cosx/x^2)`

This gives,

`= lim_(x-gtoo) (3sqrt(1-3/x^3))/(2+cosx/x^2)`

Evaluating the limits separately,

= `(lim_(x-gtoo)3sqrt(1-3/x^3))/(lim_(x->oo)2+lim_(x->oo)cosx/x^2)`

= `(3sqrt(1-3/oo^3))/(2+lim_(x->oo)cosx/x^2)`

= `3/(2+lim_(x->oo)cosx/x^2)`

Now we have to determine the `lim_(x-gtoo)cosx/x^2`

Whatever the value x is or how big it is or how small it is, cosx would be in the range of `-1lt=cosxlt=1` .

Therefore,

`lim_(x-gtoo)cosx/x^2 = 0`

(Because it will be a value between -1 and 1 divided by a very large number which is very close to zero)

Therefore,

`lim_(x-gtoo) (3sqrt(x^4-3x))/(2x^2+cosx) = 3/(2+lim_(x->oo)cosx/x^2) = 3/(2+0)`

Therefore,

`lim_(x-gtoo) (3sqrt(x^4-3x))/(2x^2+cosx) = 3/2`

The limit is `3/2` .

Use the calculator and put a very large number as 1000000000000, you will get 1.5.