Limit as x approaches infinity 4^x-(18x^2?)

I can tell whether it would be +oo or 0.

Qualitatively 4^x increases faster than x^2. So the limit of their difference as x--> oo is oo.

This shows how to work it out....


Use the Squeeze Theorem.

If Limit of g(x) = L and Lim of h(x) = L and g(x) < f(x) < h(x) then the limit of f(x) = L

The taylor series of e^x = 1 + x + x^2/2! + x^3/3! + ...

So

x^2/2 + x^3/6 - 18x^2 < e^x - 18x^2 < 4^x - 18x^2 < 4^x.

Limit 4^x -> oo as x-> oo

x^2/2 + x^3/6 - 18x^2 = x^3/6 - 35x^2/2 ==> x^3/6 > 35x^2/2 == x/6 > 35/2 ==> x > 105

So when x > 105, x^3/6 > 35x^2/2.

As x -> oo, x^3/6 ==> oo

So apply the Squeeze Theorem with g(x) = x^3/6 and h(x) = 4^x.

If x==> -infinity the limit is -infinity as 4^x==>0
If x==>+infinity 4^x -18^x^2 = e^xln4 -e^x^2ln 18 = ex^2ln18 `[e^(xln4-x^2ln18)--1 ] and as xln4-x^2ln18 ==> -infinity the brackets have lim -1 and the lim is - infinity