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Examine the following functions for continuity $\; f(x) = \large \frac {x^2 - 25} {x + 5}$$, x\:\neq\: -5$

This is a multipart question answered separately on Clay6

$\begin{array}{1 1} Yes,its\;continuous \\ No,it\;is\;not\;continuous \end{array} $

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Toolbox:
  • If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
  • Every polynomial function $f(x)$ is continuous.
Step 1:
$f(x)=\large\frac{x^2-25}{x+5}$
At $x=-5$
$f(x)=\large\frac{x^2-25}{-5+5}$
$\;\;\;\;\;\;=\large\frac{x^2-25}{0}$=Not defined.
$f$ is continuous at $x=-5$
Step 2:
At $x=c\neq -5$
$\lim\limits_{\large x\to c}f(x)=\lim\limits_{\large x\to c}\large\frac{x^2-25}{x+5}$
$\qquad\;\;\;\;\;=\lim\limits_{\large x\to c}\large\frac{x^2-(5)^2}{x+5}$
$\qquad\;\;\;\;\;=\lim\limits_{\large x\to c}\large\frac{(x-5)(x+5)}{x+5}$
$\qquad\;\;\;\;\;=\lim\limits_{\large x\to c}=c-5.$
$f(c)=c-5$
$f$ is continuous for all $x\in R-|c-5|$
answered May 27, 2013 by sreemathi.v
edited May 27, 2013 by sreemathi.v
 

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