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

From Math Wiki
Jump to navigation Jump to search
(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)  ...")
 
 
Line 10: Line 10:
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
|'''L'Hôpital's Rule'''  
+
|'''L'Hôpital's Rule, Part 1'''  
 
|-
 
|-
|&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; Let &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow c}f(x)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow c}g(x)=0,</math>&nbsp; where &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g</math>&nbsp; are differentiable functions
 
|-
 
|-
|
+
|&nbsp; &nbsp; &nbsp; &nbsp;on an open interval &nbsp;<math style="vertical-align: 0px">I</math>&nbsp; containing &nbsp;<math style="vertical-align: -5px">c,</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g'(x)\ne 0</math>&nbsp; on &nbsp;<math style="vertical-align: 0px">I</math>&nbsp; except possibly at &nbsp;<math style="vertical-align: 0px">c.</math>&nbsp;
&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: -18px">\lim_{x\rightarrow c} \frac{f(x)}{g(x)}=\lim_{x\rightarrow c} \frac{f'(x)}{g'(x)}.</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>
 
 
|}
 
|}
  
Line 44: Line 43:
 
|-
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
|&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}}} & = & \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})}{9-(9-x)}}\\
 
&&\\
 
&&\\
& = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9+x})}{-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{\lim_{x\rightarrow 0} \frac{3+\sqrt{9-x}}{1}}\\
 
&&\\
 
&&\\
& = & \displaystyle{ \frac{3+\sqrt{9}}{-1}}\\
+
& = & \displaystyle{ \frac{3+\sqrt{9}}{1}}\\
 
&&\\
 
&&\\
& = & \displaystyle{-\frac{6}{1}}\\
+
& = & \displaystyle{\frac{6}{1}}\\
 
&&\\
 
&&\\
& = & \displaystyle{-6.}
+
& = & \displaystyle{6.}
 
\end{array}</math>
 
\end{array}</math>
 
|-
 
|-
Line 121: Line 120:
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;<math>-6</math>
+
|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;<math>6</math>
 
|-
 
|-
 
|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;<math>1</math>
 
|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;<math>1</math>

Latest revision as of 18:14, 20 May 2017

Compute

(a)  

(b)  

(c)  

Foundations:  
L'Hôpital's Rule, Part 1

        Let    and    where    and    are differentiable functions

       on an open interval    containing    and    on    except possibly at   
       Then,  


Solution:

(a)

Step 1:  
We begin by noticing that we plug in    into
       
we get  
Step 2:  
Now, we multiply the numerator and denominator by the conjugate of the denominator.
Hence, we have
       

(b)

Step 1:  
We proceed using L'Hôpital's Rule. So, we have

       

Step 2:  
Now, we plug in    to get
       

(c)

Step 1:  
We begin by factoring the numerator and denominator. We have

       

So, we can cancel    in the numerator and denominator. Thus, we have

       

Step 2:  
Now, we can just plug in    to get
       


Final Answer:  
   (a)   
   (b)   
   (c)   

Return to Sample Exam