# Prove that the limit of a polynomial from the right and left exists, but the limit of the polynomial does not exists.

Suppose that P is a polynomial and that P(a) > 0 for a fixed real value a. Prove that P(x)/ (x-a) approaches infinity as x approaches a^{+}, P(x)/(x-a) approaches negative infinity as x approaches a^{-} , but limit of P(x)/(x-a) does not exist.

Note: I have to prove it the Real Analysis way. Using the Definitions. I will try to use this definitions. Definition 3.12: Let a \epsilon R and f a real function. 1) f(x) is said to converge to L as x approaches a from the right if and only if f is defined on some open interval I with left endpoints a and for every \epsilon > 0 there is a \delta > 0 such that a < x < a + \delta imlies |f(x) - L| < \epsilon. In this case we call L the right hand limit of f at a. Similarly is the left hand limit.

Definition when converging to infinity. 2) The function f(x) is said to converge to infinity as x approaches a if and only if there is an open interval I containing a such that given a real M, there is an delta > 0 such that 0 < |x - a| < delta implies f(x) > M, in which case we shall write f(x) approaches infinity as x approaches a. Similarly when converging to negative infinity.

So I need to mention delta, epsilon and M.

I am having stuck. Please can someone please help me? I would really appreciate it. Thank you. I think we can use if P is a polynomial, then the limit of P(x) as x approaches a is equal to P(a).

## 1 Answer

Outline: Let $P(a)=b\gt 0$. Then by the continuity of $P(x)$ at $x=a$, there exists a $\delta_1$ such that if $|x-a|\lt \delta_1$, then $|P(x)-b|\lt \frac{b}{2}$. In particular, if $|x-a|\lt \delta_1$, we have $P(x)\gt \frac{b}{2}$.

We want to show that for any $M\gt 0$, there exists a $\delta$ such that if $0\lt |x-a|\lt \delta$, then $\left|\frac{P(x)}{x-a}\right|\gt M$.

We will make sure later that $\delta\le \delta_1$. Then if $0\lt |x-a|\lt \delta$, we have $\frac{P(x)}{|x-a|}\gt \frac{b}{2\delta}$. To make this $\gt M$, it is enough to take $\delta\le \frac{b}{2M}$.

Putting things together, we want $\delta=\min\left(\delta_1, \frac{b}{2M}\right)$.

If $0\lt x-a\lt \delta$, then $\frac{P(x)}{x-a}$ is positive, and greater than $M$. It follows that $$\lim_{x\to a^+}\frac{P(x)}{x-a}=\infty.$$ By the same inequalities, if $x\lt a$ and $|x-a|\lt \delta$, then $\frac{P(x)}{x-a}$ is less than $-M$. It follows that $$\lim_{x\to a^-}\frac{P(x)}{x-a}=-\infty.$$

• Thank you very much for your help. It really helps. – user3492 Oct 4 '14 at 2:58
• You are welcome. I wrote it rather austerely, I hope the motivation is clear. We want to make sure that $P(x)$ doesn't get too small. Then if $|x-a|$ is tiny, the ratio is very big on the right, big negative on the left. – André Nicolas Oct 4 '14 at 3:02