009A Sample Final 3, Problem 7 Detailed Solution
Revision as of 15:45, 2 December 2017 by Kayla Murray (talk | contribs) (Created page with "<span class="exam">Compute <span class="exam">(a) <math style="vertical-align: -18px">\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}</math> <span class="exam">(b) ...")
Compute
(a)
(b)
(c)
| Background Information: |
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| L'Hôpital's Rule, Part 1 |
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Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow c}f(x)=0} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow c}g(x)=0,} where and are differentiable functions |
| on an open interval containing Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle c,} and on except possibly at |
| Then, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow c}{\frac {f(x)}{g(x)}}=\lim _{x\rightarrow c}{\frac {f'(x)}{g'(x)}}.} |
Solution:
(a)
| Step 1: |
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| We begin by noticing that we plug in into |
| we get |
| Step 2: |
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| Now, we multiply the numerator and denominator by the conjugate of the denominator. |
| Hence, we have |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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}}} |
(b)
| Step 1: |
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| We proceed using L'Hôpital's Rule. So, we have |
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|
| Step 2: |
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| Now, we plug in to get |
(c)
| Step 1: |
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| We begin by factoring the numerator and denominator. We have |
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| So, we can cancel Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x+2} in the numerator and denominator. Thus, we have |
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| Step 2: |
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| Now, we can just plug in 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 x=-2} to get |
| 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\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}} |
| Final Answer: |
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| (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 6} |
| (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 1} |
| (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 -\frac{5}{12}} |