Difference between revisions of "009C Sample Final 1, Problem 1"

Revision as of 18:34, 8 February 2016

Compute

a) $\lim _{n\rightarrow \infty }{\frac {3-2n^{2}}{5n^{2}+n+1}}$

b) $\lim _{n\rightarrow \infty }{\frac {\ln n}{\ln 3n}}$

Foundations:
Review L'Hopital's Rule

Solution:

(a)

Step 1:
First, we switch to the limit to $x$ so that we can use L'Hopital's rule.
So, 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 \infty }{\frac {3-2x^{2}}{5x^{2}+x+1}}}&{\overset {l'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {-4x}{10x+1}}}\\&&\\&{\overset {l'H}{=}}&\displaystyle {\frac {-4}{10}}\\&&\\&=&\displaystyle {\frac {-2}{5}}\end{array}}}

Step 2:
Hence, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{n\rightarrow \infty }{\frac {3-2n^{2}}{5n^{2}+n+1}}={\frac {-2}{5}}} .

(b)

Step 1:
Again, we switch to the limit to $x$ so that we can use L'Hopital's rule.
So, we have
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 \infty}\frac{\ln x}{\ln(3x)}} & \overset{l'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{\frac{1}{x}}{\frac{3}{3x}}}\\ &&\\ & = & \displaystyle{\lim_{x \rightarrow \infty} 1}\\ &&\\ & = & 1 \end{array}}

Step 2:
Hence, we have
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_{n\rightarrow \infty} \frac{\ln n}{\ln 3n}=1} .

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 \frac{-2}{5}}
(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}

@@ Line 37: / Line 37: @@
 |Hence, we have
 |-
-|<math>\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}=\frac{-2}{5}</math>
+|<math>\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}=\frac{-2}{5}</math>.
-|-
-|
 |}
@@ Line 47: / Line 45: @@
 !Step 1: &nbsp;
 |-
-|
+|Again, we switch to the limit to <math>x</math> so that we can use L'Hopital's rule.
 |-
-|
+|So, we have
 |-
 |
+::<math>\begin{array}{rcl}
+\displaystyle{\lim_{x \rightarrow \infty}\frac{\ln x}{\ln(3x)}} & \overset{l'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{\frac{1}{x}}{\frac{3}{3x}}}\\
+&&\\
+& = & \displaystyle{\lim_{x \rightarrow \infty} 1}\\
+&&\\
+& = & 1
+\end{array}</math>
 |}
@@ Line 57: / Line 62: @@
 !Step 2: &nbsp;
 |-
-|
+|Hence, we have
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 3: &nbsp;
 |-
-|
+|<math>\lim_{n\rightarrow \infty} \frac{\ln n}{\ln 3n}=1</math>.
-|-
-|
 |}
@@ Line 73: / Line 72: @@
 |'''(a)''' <math>\frac{-2}{5}</math>
 |-
-|'''(b)'''
+|'''(b)''' <math>1</math>
 |}
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]