Menu

Prove that $\lim_{x\to 3}\,f(x)=9$ where $f(x)=\left\{\begin{array}{c}3x^2-18&x\neq 3 \\ 5 & x=3\\\end{array}\right.$

Primary tabs

Lesson Parent:

Problem:

Prove that
$\lim_{x\to 3}\,f(x)=9$ where
$f(x)=\left\{\begin{array}{x}3x^2-18&x\neq 3 \\ 5 & x=3\\\end{array}\right.$

Answer:

It is true that $\lim_{x\to 3}\,f(x)=9$ where $f(x)=\left\{\begin{array}{x}3x^2-18&x\neq 3 \\ 5 & x=3\\\end{array}\right.$

Solution:

This is one of those proofs to show that a discontinuity in a function does not affect the value of a function limit. If the function were just <Sign in to see all the formulas> then at $x=3$ <Sign in to see all the formulas>. However, at $x=3$ , <Sign in to see all the formulas>.

But this definition does not require $x$ to be equal to $3$ . We know this because the definition says "if <Sign in to see all the formulas>" which means that <Sign in to see all the formulas>. Since, we assume that we are at <Sign in to see all the formulas>, then our <Sign in to see all the formulas> and we already know that <Sign in to see all the formulas> because

The proof is complete.