022 Exam 1 Sample A, Problem 3

Problem 3. Given a function $g(x)={\frac {x+5}{x^{2}-25}}$ ,

(a) Find the intervals where $g(x)$ is continuous.

(b). Find $\lim _{x\rightarrow 5}g(x)$.

Foundations:
A function $f$ is continuous at a point $x_{0}$ if
$\lim _{x\rightarrow x_{_{0}}}f(x)=f\left(x_{0}\right).$
This can be viewed as saying the left and right hand limits exist, and are equal to the value of $f$ at $x_{0}$.

 Solution:

(a):
Note that
$g(x)\,\,=\,\,{\frac {x+5}{x^{2}-5}}\,\,=\,\,{\frac {x+5}{(x-5)(x+5)}}.$
In order to be continuous at a point $x_{0}$, $f(x_{0})$ must exist. However, attempting to plug in $x=\pm 5$ results in division by zero. Therefore, in interval notation, we have that $f$ is continuous on
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-\infty ,-5)\cup (-5,5)\cup (5,\infty). }

(b):
Note that in order for the limit to exist, the limit from both the left and the right must be equal. But
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} {\displaystyle \lim_{x\rightarrow5^{+}}g(x)} & = & {\displaystyle \lim_{x\rightarrow5^{+}}\frac{x+5}{(x-5)(x+5)}}\\ \\ & = & {\displaystyle \lim_{x\rightarrow5^{+}}\frac{1}{x-5}}\\ \\ & = & {\displaystyle \frac{1}{\,\,0^{+}}\rightarrow+\infty,} \end{array}}
while
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} {\displaystyle \lim_{x\rightarrow5^{-}}g(x)} & = & {\displaystyle \lim_{x\rightarrow5^{-}}\frac{x+5}{(x-5)(x+5)}}\\ \\ & = & {\displaystyle \lim_{x\rightarrow5^{-}}\frac{1}{x-5}}\\ \\ & = & {\displaystyle \frac{1}{\,\,0^{-}}\rightarrow-\infty,} \end{array}}
where  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0^{-}} can be thought of as "really small negative numbers approaching zero." Since the handed limits do not agree, the limit as x approaches 5 does not exist.

Final Answer:
(a): Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f } is continuous on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-\infty ,-5)\cup (-5,5)\cup (5,\infty). }
(b): Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow5}g(x)} does not exist.

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