Difference between revisions of "009A Sample Final 3, Problem 7"

Latest revision as of 16:44, 2 December 2017

Compute

(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 \lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}}

(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\rightarrow \pi} \frac{\sin x}{\pi-x}}

(c) 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\rightarrow -2} \frac{x^2-x-6}{x^3+8}}

Solution

Detailed Solution

Return to Sample Exam

@@ Line 7: / Line 7: @@
 <span class="exam">(c) &nbsp;<math style="vertical-align: -16px">\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009A Sample Final 3, Problem 7 Solution|'''<u>Solution</u>''']]
-|-
-|'''L'Hôpital's Rule'''
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; Suppose that &nbsp;<math style="vertical-align: -11px">\lim_{x\rightarrow \infty} f(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -11px">\lim_{x\rightarrow \infty} g(x)</math>&nbsp; are both zero or both &nbsp;<math style="vertical-align: -1px">\pm \infty .</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}</math>&nbsp; is finite or &nbsp;<math style="vertical-align: -4px">\pm \infty ,</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; then &nbsp;<math style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}\,=\,\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
-|}
-'''Solution:'''
+[[009A Sample Final 3, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']]
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We begin by noticing that we plug in &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; into
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{x}{3-\sqrt{9-x}},</math>
-|-
-|we get &nbsp; <math style="vertical-align: -12px">\frac{0}{0}.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, we multiply the numerator and denominator by the conjugate of the denominator.
-|-
-|Hence, we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}} & = & \displaystyle{\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}\frac{(3+\sqrt{9+x})}{(3+\sqrt{9+x})}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9+x})}{9-(9+x)}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9+x})}{-x}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow 0} \frac{3+\sqrt{9+x}}{-1}}\\
-&&\\
-& = & \displaystyle{ \frac{3+\sqrt{9}}{-1}}\\
-&&\\
-& = & \displaystyle{-\frac{6}{1}}\\
-&&\\
-& = & \displaystyle{-6.}
-\end{array}</math>
-|-
-|
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We proceed using L'Hôpital's Rule. So, we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{\lim_{x\rightarrow \pi} \frac{\sin (x)}{\pi-x}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \pi}\frac{\cos(x)}{-1}.}
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, we plug in &nbsp;<math style="vertical-align: 0px">x=\pi</math>&nbsp; to get
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{x\rightarrow \pi} \frac{\sin (x)}{\pi-x}} & = & \displaystyle{\frac{\cos(\pi)}{-1}}\\
-&&\\
-& = & \displaystyle{\frac{-1}{-1}}\\
-&&\\
-& = & \displaystyle{1.}
-\end{array}</math>
-|}
-'''(c)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We begin by factoring the numerator and denominator. We have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}\,=\,\lim_{x\rightarrow -2}\frac{(x+2)(x-3)}{(x+2)(x^2-2x+4)}.</math>
-|-
-|So, we can cancel &nbsp;<math style="vertical-align: -2px">x+2</math>&nbsp; in the numerator and denominator. Thus, we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}\,=\,\lim_{x\rightarrow -2}\frac{x-3}{x^2-2x+4}.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, we can just plug in &nbsp;<math style="vertical-align: -1px">x=-2</math>&nbsp; to get
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}} & = & \displaystyle{\frac{-2-3}{(-2)^2-2(-2)+4}}\\
-&&\\
-& = & \displaystyle{-\frac{5}{12}.}
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;<math>-6</math>
-|-
-|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;<math>1</math>
-|-
-|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp;<math>-\frac{5}{12}</math>
-|}
 [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009A Sample Final 3, Problem 7"

Latest revision as of 16:44, 2 December 2017

Navigation menu

Search