022 Sample Final A, Problem 14 - Revision history

MathAdmin at 00:19, 7 June 2015

2015-06-07T00:19:28Z

MathAdmin at 21:03, 6 June 2015

2015-06-06T21:03:41Z

MathAdmin: Created page with " Find the following limit: $\qquad \lim_{x \rightarrow \,-3}\frac{x^2 + 7x + 12}{x^2 - 2x - 14}$ . {| class="mw-col..."

2015-06-05T02:59:36Z

Created page with " Find the following limit: <math style="vertical-align: -15px">\qquad \lim_{x \rightarrow \,-3}\frac{x^2 + 7x + 12}{x^2 - 2x - 14}</math>. {| class="mw-col..."

New page

Find the following limit: <math style="vertical-align: -15px">\qquad \lim_{x \rightarrow \,-3}\frac{x^2 + 7x + 12}{x^2 - 2x - 14}</math>.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|When evaluating limits of rational functions, the first idea to try is to simply plug in the limit. In addition to this, we must consider that as a limit,
|-
|     <math>\frac {1}{\infty} = 0,</math>
|-
|and
|-
|     <math>\lim_{x\rightarrow 0^{\pm}}\frac{1}{x}\,=\,\pm \infty.</math>
|-
|In the latter case, the sign matters. Unfortunately, most (but not all) exam questions require more work. Many of them will evaluate to an '''indeterminate form''', or something of the form
|-
|      <math>\frac{0}{0}</math>   or   <math>\frac {\pm \infty}{\pm \infty}.</math>
|-
| In this case, there are several approaches to try:
|-
|
*We can multiply the numerator and denominator by the conjugate of the denominator. This frequently results in a term that cancels, allowing us to then just plug in our limit value.
*We can factor a term creatively. For example, <math style="vertical-align: -9%;">x-1</math> can be factored as <math style="vertical-align: -48%;">\left(\sqrt{x}-1\right) \left(\sqrt{x}+1\right)</math> , or as <math style="vertical-align: -71%;">\left(\sqrt[3]{x}-1\right)\left(\left(\sqrt[3]{x}\right)^{2}+\sqrt[3]{x}+1\right)</math> , both of which could result in a factor that cancels in our fraction.
*We can apply '''l'Hôpital's Rule:''' Suppose <math style="vertical-align: 0%;">c</math> is contained in some interval <math style="vertical-align: 0%;">I</math>. If <math style="vertical-align: -60%;">\lim_{x \to c}f(x)=\lim_{x \to c}g(x)=0 \text{ or } \pm\infty</math>  and <math style="vertical-align: -85%;">\lim_{x\to c}\frac{f'(x)}{g'(x)}</math>   exists, and <math style="vertical-align: -24%;">g'(x)\neq 0</math>  for all <math style="vertical-align: -23%;">x\neq c</math>  in <math style="vertical-align: 0%;">I</math>, then <math style="vertical-align: -84%;">\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}</math>.
|-
|Note that the first requirement in l'Hôpital's Rule is that the fraction ''must'' be an indeterminate form. This should be shown in your answer for any exam question. 
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution: 
|-
|Usually, we first want to try simply plugging in the limit value. In this case, we find that
|-
|
::<math>\frac{(-3)^2 + 7(-3) + 12}{(-3)^2 - 2(-3) - 14}\,=\,\frac{9-21+12}{9+6-14}\,=\,\frac{0}{1}\,=\,0.</math>
|-
|Since we can just plug in the value, the limit is <math style="vertical-align: 0px">0</math>.

|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|
::<math style="vertical-align: -15px">\qquad \lim_{x \rightarrow \,-3}\frac{x^2 + 7x + 12}{x^2 - 2x - 14}\,=\,0.</math>
|}

[[022_Sample_Final_A|'''Return to Sample Exam''']]

@@ Line 34: / Line 34: @@
 ::<math>\lim_{x \rightarrow -3}\frac{(x)^2 + 7(x) + 12}{(x)^2 - 2(x) - 15}\,=\,\frac{9-21+12}{9+6-15}\,=\,\frac{0}{0}.</math>
 |-
-|Thus, we need to Apply L'hopital's Rule.
+|This is an indeterminate form, and we need to apply l'Hôpital's Rule.
 |}
@@ Line 40: / Line 40: @@
 !Step 2:&nbsp;
 |-
-|After applying L'hopital's Rule we find:
+|Applying l'Hôpital's Rule (by taking the derivative of the numerator and denominator separately), we find:
 |-
+|
 ::<math>\begin{array}{rcl}
-\displaystyle{\lim_{x \rightarrow -3}\frac{x^2 + 7x + 12}{x^2 - 5x -15}} & = & \displaystyle{\lim_{x \rightarrow -2}\frac{2x + 7}{2x -2}}\\
+\displaystyle{\lim_{x \rightarrow -3}\frac{x^2 + 7x + 12}{x^2 - 5x -15}} & \overset{l'H}{=} & \displaystyle{\lim_{x \rightarrow -3}\frac{2x + 7}{2x -2}}\\
 &&\\
 & = & \displaystyle{\frac{2(-3) + 7}{2(-3) - 2}}\\
 &&\\
-& = & \displaystyle{\frac{1}{-8}}
+& = & \displaystyle{\frac{~1}{-8}.}
 \end{array}</math>
 |}

@@ Line 1: / Line 1: @@
-<span class="exam"> Find the following limit: <math style="vertical-align: -15px">\qquad \lim_{x \rightarrow \,-3}\frac{x^2 + 7x + 12}{x^2 - 2x - 14}</math>.
+<span class="exam"> Find the following limit: <math style="vertical-align: -15px">\qquad \lim_{x \rightarrow \,-3}\frac{x^2 + 7x + 12}{x^2 - 2x - 15}</math>.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 25: / Line 25: @@
 |Note that the first requirement in l'H&ocirc;pital's Rule is that the fraction <u>''must''</u> be an indeterminate form.  This should be shown in your answer for any exam question.<br>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Solution:&nbsp;
+!Step 1:&nbsp;
 |-
-|Usually, we first want to try simply plugging in the limit value.  In this case, we find that
+|We take the limit and find that
 |-
+|
-::<math>\frac{(-3)^2 + 7(-3) + 12}{(-3)^2 - 2(-3) - 14}\,=\,\frac{9-21+12}{9+6-14}\,=\,\frac{0}{1}\,=\,0.</math>
+::<math>\begin{array}{rcl}
-|-
+\displaystyle{\lim_{x \rightarrow -3}\frac{x^2 + 7x + 12}{x^2 - 5x -15}} & = & \displaystyle{\lim_{x \rightarrow -2}\frac{2x + 7}{2x -2}}\\
-|Since we can just plug in the value, the limit is <math style="vertical-align: 0px">0</math>.
+&&\\
+& = & \displaystyle{\frac{2(-3) + 7}{2(-3) - 2}}\\
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 43: / Line 56: @@
 |-
+|
-::<math style="vertical-align: -15px">\qquad \lim_{x \rightarrow \,-3}\frac{x^2 + 7x + 12}{x^2 - 2x - 14}\,=\,0.</math>
+::<math style="vertical-align: -15px">\qquad \lim_{x \rightarrow \,-3}\frac{x^2 + 7x + 12}{x^2 - 2x - 14}\,=\, -\frac{1}{8}.</math>
 |}
 [[022_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

022 Sample Final A, Problem 14 - Revision history

MathAdmin at 00:19, 7 June 2015

MathAdmin at 21:03, 6 June 2015

MathAdmin: Created page with " Find the following limit: \qquad \lim_{x \rightarrow \,-3}\frac{x^2 + 7x + 12}{x^2 - 2x - 14}. {| class="mw-col..."

MathAdmin: Created page with " Find the following limit: $\qquad \lim_{x \rightarrow \,-3}\frac{x^2 + 7x + 12}{x^2 - 2x - 14}$ . {| class="mw-col..."