022 Exam 1 Sample A, Problem 3 - Revision history

MathAdmin at 18:06, 12 April 2015

2015-04-12T18:06:57Z

MathAdmin at 18:01, 12 April 2015

2015-04-12T18:01:34Z

MathAdmin: Created page with "'''Problem 3.''' Given a function $g(x)=\frac{x+5}{x^{2}-25}$ , :
2015-04-12T17:54:53Z

Created page with "<span class="exam">'''Problem 3.''' Given a function <math style="vertical-align: -40%;">g(x)=\frac{x+5}{x^{2}-25}</math> , :<span..."

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<span class="exam">[[022_Exam_1_Sample_A,_Problem_3 |'''Problem 3.''']] Given a function <math style="vertical-align: -40%;">g(x)=\frac{x+5}{x^{2}-25}</math> ,

:<span class="exam">(a) Find the intervals where <math style="vertical-align: -20%;">g(x)</math> is continuous.

:<span class="exam">(b). Find <math style="vertical-align: -55%;">\lim_{x\rightarrow5}g(x)</math>.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Foundations:  
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|A function <math style="vertical-align: -20%;">f</math> is continuous at a point <math style="vertical-align: -14%">x_0 </math> if
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::<math>\lim_{x\rightarrow x_{_0}} f(x) = f\left(x_0\right).</math>
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|This can be viewed as saying the left and right hand limits exist, and are equal to the value of <math style="vertical-align: -20%">f</math> at <math style="vertical-align: -14%">x_0</math>.<br>
|}

 '''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! (a):  
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|Note that
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::<math>g(x)\,\,=\,\,\frac{x+5}{x^2-5}\,\,=\,\,\frac{x+5}{(x-5)(x+5)}.</math>
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|In order to be continuous at a point <math style="vertical-align: -14%">x_0 </math>, <math style="vertical-align: -20%">f(x_0) </math> must exist. However, attempting to plug in <math style="vertical-align: 0%">x=\pm 5 </math> results in division by zero. Therefore, in interval notation, we have that <math style="vertical-align: -20%">f </math> is continuous on
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::<math>(-\infty ,-5)\cup (-5,5)\cup (5,\infty). </math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! (b):  
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|Note that in order for the limit to exist, the limit from both the left and the right must be equal. But
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::<math>\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}</math>
|-
|while
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::<math>\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}</math>
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|where  <math style="vertical-align: 0%">0^{-}</math> 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.
|}
[[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]