Chapter 12, Problem 41CLT

To determine

Toevaluate:Thelimit $L=\underset{x\to \infty }{\lim }\frac{3x^2+1}{x^2+4}$ and if exists verify the results using graphing utility.

If limit does not exist explain why.

Answer to Problem 41CLT

The limit L exists and $L=\underset{x\to \infty }{\lim }\frac{3x^2+1}{x^2+4}=3$ .

Explanation of Solution

Given information:

The given limit is $L=\underset{x\to \infty }{\lim }\frac{3x^2+1}{x^2+4}$

Formulaused:

Let f be a function $f:R\to R$  is defined on the real line and  a, L  ∈  R . Then L is said the limit of function f , as  x  approaches  a ,and is written as $\underset{x\to a}{\lim }f\left(x\right)=L$ .The limit $\underset{x\to a}{\lim }f\left(x\right)=L$ is calculated by solving for finite value of $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)=L$ by manipulating to exist Or by calculating $\underset{h\to 0}{\lim }f\left(a+h\right)=L=\underset{h\to 0}{\lim }f\left(a-h\right)$

Calculations:

The given limit is $L=\underset{x\to \infty }{\lim }\frac{3x^2+1}{x^2+4}$

  $L=\underset{x\to \infty }{\lim }\frac{\frac{3x^2+1}{x^2}}{\frac{x^2+4}{x^2}}$

  $L=\underset{x\to \infty }{\lim }\frac{3+\frac{1}{x^2}}{1+\frac{4}{x^2}}$

  $\Rightarrow L=\frac{3+\frac{1}{{\infty }^2}}{1+\frac{4}{{\infty }^2}}=\frac{3+0}{1+0}$

  $\Rightarrow L=3$

Thus, the limit L exists and $L=3$

Verification of limit with graph:

The graph of function $f\left(x\right)=\frac{3x^2+1}{x^2+4}$ is shown in Figure-1.

  

From the graph it is clear that the limit $L=\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{3x^2+1}{x^2+4}=3$ .

Conclusion:

The limit Lexists and $L=\underset{x\to \infty }{\lim }\frac{3x^2+1}{x^2+4}=3$ .