Q. 12 - Calculus

Question

Fill in the blanks to complete each of the following theorem statements: 

If $\underset{x\to c}{\lim }g\left(x\right)$ exists and $f\left(x\right)$ is a function that is ? to $g\left(x\right)$ for all $x$ sufficiently close to ?, but not necessarily at ? , then  ?.

Step-by-Step Solution

Answer

If $\underset{x\to c}{\lim }g\left(x\right)$ exists and $f\left(x\right)$is a function that is equal to $g$ for all $x$ sufficiently close to $c$, but not necessarily at $c$, then $\underset{x\to c}{\lim }f\left(x\right)=\underset{x\to c}{\lim }g\left(x\right).$

Step 1. Given Information.

The given statement is:

If $\underset{x\to \infty }{\lim }g\left(x\right)$ exists and $f\left(x\right)$is a function that is equal to ? for all $x$ sufficiently close to ? but not necessarily at ?, then ?

Step 2. Fill the blanks.

According to the Cancellation Theorem for Limits,

If $\underset{x\to \infty }{\lim }g\left(x\right)$, and $f$ is a function that is equal to $g$ for all $x$ sufficiently close to $c$ except possibly at $c$ itself, then $\underset{x\to c}{\lim }f\left(x\right)=\underset{x\to c}{\lim }g\left(x\right).$