Q. 85 - Calculus

Question

Use limits to prove that the limits of a polynomial $f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+a_{1}x+a_0$ are the same as the limits of its leading term $a_n{x}^n$ as $x\to \infty$ and as $x\to -\infty$. (Hint: Show that $\underset{x\to \infty }{\lim }f\left(x\right)$ is equal to $\underset{x\to \infty }{\lim }{a}_n{x}^n$ by factoring out $a_n{x}^n$ from $f\left(x\right)$.)

Step-by-Step Solution

Answer

The limits of a polynomial $f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+a_{1}x+a_0$ are the same as the limits of its leading term $a_n{x}^n$ as $x\to \infty$ and as $x\to -\infty$.

Step 1. Given Information

We are given a function,

$f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+a_{1}x+a_0$

Step 2. Proving the statement

Take limit in the function as below,

$$\begin{gathered} \underset{x\to \infty }{\lim }\left(a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0\right)=\underset{x\to \infty }{\lim }\left(a_n{x}^n\left(1+\frac{a_{n-1}}{a_{n}x}+\cdots +\frac{a_1}{a_n{x}^{n-1}}+\frac{a_0}{a_0{x}^n}\right)\right) \\ =\left(\underset{x\to \infty }{\lim }{a}_n{x}^n\right)(1+0+\cdots +0+0) \\ =\underset{x\to \infty }{\lim }{a}_n{x}^n \end{gathered}$$

Similarly for $\underset{x\to -\infty }{\lim }f\left(x\right)$.

Hence, limits of a polynomial $f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+a_{1}x+a_0$ are the same as the limits of its leading term $a_n{x}^n$ as $x\to \infty$ and as $x\to -\infty$.