Let $f(x) = x^3-x^2–3x–1$ and $h(x)=\frac{f(x)}{g(x)}$ where $h(x)$ is a rational function such that

Let $f(x) = x^3-x^2–3x–1$ and $h(x)=\dfrac{f(x)}{g(x)}$ where $h(x)$ is a rational function such that

(a) it is continuous every where except when $x=– 1$

(b) $\lim_{x\to\infty}=\infty$

(c) $\lim_{x\to-1}=\dfrac{1}{2}$

Find $\lim_{x\to0}(3h(x)+f(x)-2g(x))$

My attempt is as follows:-

$$\lim_{x\to-1}h(x)=\lim_{x\to-1}\dfrac{f(x)}{g(x)}$$

As $x\rightarrow -1$, $f(x)\rightarrow 0$

So $g(x)$ should also tend to zero as $x\rightarrow -1$, otherwise R.H.S will be equal to $0$ which is not equal to L.H.S. $$\lim_{x\to-1}g(x)=0\tag{1}$$

Applying L' Hospital rule

$$\lim_{x\to-1}\dfrac{3x^2-2x-3}{g'(x)}=\dfrac{1}{2}$$ $$\dfrac{\lim_{x\to-1}(3x^2-2x-3)}{\lim_{x\to-1}g'(x)}=\dfrac{1}{2}$$ $$\lim_{x\to-1}g'(x)=4\tag{2}$$

$$\lim_{x\to\infty}h(x)=\lim_{x\to\infty}\dfrac{f(x)}{g(x)}$$

As $x\rightarrow \infty$, $f(x)\rightarrow \infty$ because $x^3$ will be a very huge term in comparison to other lower degree terms when $x\rightarrow \infty$

Can we say anything about $\lim_{x\to\infty}g(x)$? I am not able to draw any conclusion for $\lim_{x\to\infty}g(x)$

Let's find following with the information we derived

$$\lim_{x\to0}(3h(x)+f(x)-2g(x))$$ $$\lim_{x\to0}\left(3\dfrac{f(x)}{g(x)}+f(x)-2g(x)\right)=\dfrac{-3}{\lim_{x\to0}g(x)}-1-2\lim_{x\to0}g(x)$$

I am stuck here, not getting any ideas to how to proceed further. Any hints?

2 Answers

If we have that $g$ is a polynomial, we know the following things:

$(1)$ Since $h$ is continuous everywhere except $-1$, $g$ can only have a zero at $-1$, hence $g$ is of the form $a(x+1)^n$.

$(2)$ Since $h\to\infty$ as $x\to\infty$, that means that $a>0$ and $g$ must be a degree $2$ or $1$ polynomial because the cube in numerator has to dominate.

$(3)$ Factoring $f$, we see that $f(x) = (x+1)(x^2-2x-1)$. Since the limit as $x\to -1$ of $h$ is finite, $g$ can only have a zero of multiplicity $1$ at $-1$. This means $n=1$. Plugging in we can also solve for $a$, which is $$\frac{(-1)^2+2-1}{\frac{1}{2}} = 4$$

This gives us that $g(x)=4x+4$. Then the limit we want is $-\frac{39}{4}$

(a) tells us that $g(-1)=0$ and is the only real root of $g(x)$.
(b) tells us that $g(x)$ less than $3$.
(c) tells us that $f(x)$ and $g(x)$ have $x+1$ as a common factor, and furthermore, $g(x)$ has only one factor of $x+1$.

From this, we determine that $g(x)$ is a linear function which can be written in the form $g(x)=k(x+1)$.

We therefore have $$\lim_{x\to-1}h(x)=\frac12=\lim_{x\to-1}\frac{(x+1)(x^2-2x-1)}{k(x+1)}\implies\frac12=\frac2k\implies k=4\implies g(x)=4(x+1)$$

Putting everything together into the expression that we need to evaluate, we have \begin{align} \lim_{x\to0}3h(x)+f(x)-2g(x)&=3\lim_{x\to0}h(x)+\lim_{x\to0}f(x)-2\lim_{x\to0}g(x)\\ &=3h(0)+f(0)-2g(0)\\ &=3\frac{f(0)}{g(0)}+f(0)-2g(0)\\ &=3\frac{-1}{4}+(-1)-2(4)\\ &=-\frac{39}4 \end{align}