Question

Question #4d7ec

Answer 1

`lim_(x->oo)((x^3*sin(1/x)-2x^2)/(1+3x^2))=-1/3`

Explanation:

`lim_(x->oo)((x^3*sin(1/x)-2x^2)/(1+3x^2))`
`=lim_(x->oo)((x^3*sin(1/x))/(1+3x^2))-lim_(x->oo)((2x^2)/(1+3x^2))`
`=lim_(x->oo)((-xcos(1/x)+3x^2*sin(1/x))/(6x))-2/3`
`=-1/6-2/3+lim_(x->oo)((-cos(1/x)/x^2)/(-2/x^2))`
`=1/6-2/3+1/2`
`=-1/3`

Here's how it works:

1. Separate a constant value from the numerator. (line 2)

2. Apply the *L'Hopital's rule to the indefinite form (line 3) (Hope you've learned this at school:)

3. Since `lim_(x->oo)(1/x)=0`, `cos(1/x)=1`