Proving that $\lim_{n\to\infty}x_n = x$ implies that $\lim_{n\to\infty}p(x_n) = p(x)$ where $p$ is a polynomial.

Question

I apologize in advance for the Length of the post.

Could you please critique the following proof and let me know whether my attempt is correct?

Proposition. Let $(x_n)\to L$ and $p(x)$ be a polynomial. Prove that $p(x_n)\to p(L)$.

The core of my argument is encapulated in the following Lemma. Given this my remaining idea was to use to use the algebraic properties of limits to $p(x_n)\to p(L)$.

---

Lemma. Given any arbitrary positive integer $r$. The sequence $(x_n^r)$ converges to $L^r$.

Proof. We construct the proof by recourse to Mathematical Induction. For $r = 1$ we are requried to show that $(x_n)\to L$ which ofcourse is true from hypothesis, establishing the basis of induction.

Now assume for an arbitrary $k\in\mathbf{Z}^+$ that $(x_n^k)\to L^k$. In addition let $M>0$ be a bound for $(x_n)$ the existence of which is guarenteed since $(x_n)$ is convergent.

Now Let $\epsilon>0$. We proceed by arguing from cases where $L=0$ and $L\neq 0$.

Adressing the former case first. By inductive hypothesis we have some $N_0\in\mathbf{Z}^+$ such that

$$\forall n\ge N_0\left(|x_n^k|<\frac{\epsilon}{M}\right)$$ but $|x_n|\leq M$ consequently $$|x_n|\cdot|x_n^{k}| = |x_n^{k+1}|<|x_n|\cdot\frac{\epsilon}{M}\leq\epsilon$$ thus $|x_n^{k+1}|<\epsilon$.

Addressing the later claim, again by inductive hypothesis and our proof of the base case. We have $N_1,N_2\in\mathbf{Z}^+$ such that $$\forall n\ge N_1\left(|x_n^k-L^{k}|<\frac{\epsilon}{2\cdot M}\right)\text{ }\text{ and }\text{ }\forall n\ge N_2\left(|x_n-L|<\frac{\epsilon}{2\cdot|L^k|}\right)$$

So for any $n\ge \max(N_1,N_2)$ we have

$$|x_n|\cdot|x_n^{k}-L^k| = |x_n^{k+1}-L^kx_n|<|x_n|\cdot\frac{\epsilon}{2\cdot M}\leq\frac{\epsilon}{2}$$

and similarly we have

$$|L^k|\cdot|x_n-L| = |L^kx_n-L^{k+1}|<|L^k|\cdot\frac{\epsilon}{2\cdot |L^k|}\leq\frac{\epsilon}{2}$$

The use of the triangle inequality then implies that $$|x^{k+1}-L^{k+1}| = |x^{k+1}-L^kx_n+L^kx_n-L^{k+1}| = |x^{k+1}-L^kx_n|+|L^kx_n-L^{k+1}|<\frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$$

$\blacksquare$

---

Answer 1

What you did is correct. But you wrote about using the algebraic properties of limits in order to prove what you want to prove about polynomials. Note that using these properties, you can get a shorter proof of your lemma:

• The case $r=1$ is trivial.
• If the statment holds for a $r$, then\begin{align}\lim_{n\to\infty}{x_n}^{r+1}&=\lim_{n\to\infty}{x_n}^r.x_n\\&=\left(\lim_{n\to\infty}{x_n}^r\right)\left(\lim_{n\to\infty}x_n\right)\\&=L^r.L\\&=L^{r+1}.\end{align}

Answer 2

If $f$ is continuous in $x$ and $\displaystyle \lim_{n\to\infty}x_n = x$ then we have $\displaystyle \lim_{n\to\infty}f(x_n) = f(x)$.

Since polynomial on $\mathbb{R}$ is continuous you have a conclusion.

Answer 3

Use $x^r-y^r =(x-y)\sum_{k=0}^{r-1}x^ky^{r-1-k}$ to bound $x_n^r-L^r$ in terms of $x_n-L$.