Chapter 2, Problem 1SP

To determine

To find the limit (if it exists)

  $\underset{x\to 0}{\lim }\left(1+x^2\right)$

Answer to Problem 1SP

  $\underset{x\to 0}{\lim }\left(1+x^2\right)=1$

Explanation of Solution

Given:

  $\underset{x\to 0}{\lim }\left(1+x^2\right)$

Concept Used:

Theorem on limits:

1. $\underset{x\to a}{\lim }\left[f\left(x\right)g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)\underset{x\to a}{\lim }g\left(x\right)$, product rule
2. $\underset{x\to a}{\lim }\left[f\left(x\right)\pm g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)\pm \underset{x\to a}{\lim }g\left(x\right)$ , addition and subtraction rule

Calculation:

To find the limit (if it exists)

  $\underset{x\to 0}{\lim }\left(1+x^2\right)$

Use theorem of limits, as

  $\underset{x\to 0}{\lim }\left(1+x^2\right)=\underset{x\to 0}{\lim }1+\underset{x\to 0}{\lim }{x}^2,\text{ using addition rule}$

Now, upper limit can be calculated as

  $\begin{array}{c}\underset{x\to 0}{\lim }\left(1+x^2\right)=\underset{x\to 0}{\lim }1+\underset{x\to 0}{\lim }{x}^2\\ =1+0^2\\ =1\end{array}$

Hence,

  $\underset{x\to 0}{\lim }\left(1+x^2\right)=1$