# Thread: Stuck on a limit problem

1. ## Stuck on a limit problem

Hey everyone, I'm looking for help on how to solve this limit problem.

This is the information provided, sorry about the way I wrote it out.
f(x)=(x^3)-(x^2)-(3x)-1

h(x)=(f(x))/(g(x))

h is continuous except at x=-1
lim x ->infinity of h(x)= infinity
lim x-> -1 of H(x)= 1/2

Find the value of lim x->0 of (3h(x)+f(x)-2g(x))

I've tried factoring out f(x) and using it in lim x->-1 of h(x)=(f(x))/(g(x)), but I keep arriving with 0 being the numerator, which can't be right since the limit's supposed to be 1/2. Any and all help would be appreciated, thanks.

2. Originally Posted by crashen
Hey everyone, I'm looking for help on how to solve this limit problem.

This is the information provided, sorry about the way I wrote it out.
f(x)=(x^3)-(x^2)-(3x)-1

h(x)=(f(x))/(g(x))

h is continuous except at x=-1
lim x ->infinity of h(x)= infinity
lim x-> -1 of H(x)= 1/2

Find the value of lim x->0 of (3h(x)+f(x)-2g(x))

I've tried factoring out f(x) and using it in lim x->-1 of h(x)=(f(x))/(g(x)), but I keep arriving with 0 being the numerator, which can't be right since the limit's supposed to be 1/2. Any and all help would be appreciated, thanks.
Since $f(x)=x^3-x^2-3x-1=(x+1)(x^2-2x-1)\,\,but\,\,h(x)$ is not continuous at -1, we must have that $g(-1)=0$ , and as $h(x) \xrightarrow [x\to \infty] {} \infty\,\,but\,\,h(x) \xrightarrow [x\to -1] {} \frac{1}{2}$, if g(x) is a polynomial function then $g(x)=(x+1)k(x)$, with k being a polynomial function of degree $\leq 1\,\;s.t.\,\;k(-1)=4$ , and as h is continuous everywhere but -1 it must be that k has no real roots, so $k(x)=4$, and we can say that:

$h(x)=\frac{x^2-2x-1}{k(x)}=\frac{x^2-2x-1}{4}\,,\;for\,\,x\neq -1$

Now substitute stuff and evaluate your limit.

Tonio