Q. 3 - Calculus

Question

Let $p\left(x\right)$be a nonzero polynomial function. Evaluate $\underset{x\to \infty }{\lim }\frac{p(x+1)}{p\left(x\right)}.$

Step-by-Step Solution

Answer

If $p\left(x\right)=\sum _{k=0}^{\infty }{a}_k{x}^k$ then the value of $\underset{x\to \infty }{\lim }\frac{p(x+1)}{p\left(x\right)} \mathrm{is} 1.$

Step 1. Given information.

 The given expression that needs to Evaluate is $\underset{x\to \infty }{\lim }\frac{p(x+1)}{p\left(x\right)}.$

Step 2. polynomial function.

Consider a polynomial,

$$\begin{gathered} p\left(x\right)=a_0{x}^0+a_1{x}^1+a_2{x}^2+\cdots +a_n{x}^n \\ p\left(x\right)=\sum _{k=0}^{\infty }{a}_k{x}^k \end{gathered}$$

So the value of $p(x+1)$will be the following.

$p(x+1)=\sum _{k=0}^{\infty }{a}_k{\left(x+1\right)}^k$

Step 3. Value of lim x → ∞ p ( x + 1 ) p ( x ) .

Determine the value of $\underset{x\to \infty }{\lim }\frac{p(x+1)}{p\left(x\right)}.$

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{p(x+1)}{p\left(x\right)}=\underset{x\to \infty }{\lim }\frac{a_k{\left(x+1\right)}^k}{a_k{x}^k} \\ =\underset{x\to \infty }{\lim }{\left(\frac{x+1}{x}\right)}^k \\ =\underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^k \\ =1^k+0 \\ =1 \end{gathered}$$

So the value of $\underset{x\to \infty }{\lim }\frac{p(x+1)}{p\left(x\right)} \mathrm{is} 1.$