MHB Finding the exact value of a limit

[NavalMonte]

I'm having a hard time starting this problem

lim of arctan(-2x^3+3x-4x) as x approaches infinity

Any help would be appreciated

 

[Evgeny.Makarov]

Start by finding the limits of $-2x^3+3x-4x$ when $x\to\infty$ (do you mean $-2x^3+3x^2-4x$?) and $\arctan(x)$ when $x\to\pm\infty$.

 

[NavalMonte]

Start by finding the limits of $-2x^3+3x-4x$ when $x\to\infty$ (do you mean $-2x^3+3x^2-4x$?) and $\arctan(x)$ when $x\to\pm\infty$.

I'm sorry, it's actually written as:

lim arctan($-2x^3+3x-4$)
x->∞

 

[Evgeny.Makarov]

The advice is the same: see where the argument of the arctangent tends to and then see what values arctangent takes there.

 

[NavalMonte]

The advice is the same: see where the argument of the arctangent tends to and then see what values arctangent takes there.

I factored the largest factor of x from the polynomial and got:

lim $x^3$=∞
x->∞

and

lim $(-2+\dfrac{3}{x^2}-\dfrac{4}{x^2})$=-2
x->∞

Would that make the:
lim arctan (-2) =lim arctan($-2x^3+3x-4$)
x->∞...x->∞Edit: I just realized that:
lim $-2x^3+3x-4$ = -∞
x->∞

Therefore,
lim arctan($-2x^3+3x-4$)= -$\dfrac{\pi}{2}$
x->∞

Would that be correct or am I totally off base?

 

[Evgeny.Makarov]

I can only repeat what I have said: find the limit of the arctangent's argument and then find what arctangent is like near that value. You, instead, found the limit of the arctangent's argument divided by $x^3$ and substituted it into arctangent. I don't understand why you divided the argument by $x^3$.

Edit: I just saw your edit, and it is correct. You may also see this discussion on StackExchange for a similar example.