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

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!Foundations:    
 
!Foundations:    
 
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|'''L'Hôpital's Rule'''  
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|'''L'Hôpital's Rule, Part 1'''  
 
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|&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>
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&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
 
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|&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>
 
 
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|&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>
 
 
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Latest revision as of 18:10, 20 May 2017

In each part, compute the limit. If the limit is infinite, be sure to specify positive or negative infinity.

(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 factoring the numerator. We have

       

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

       

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

       

(b)

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

       

Step 2:  
This limit is  

(c)

Step 1:  
We have

       

Since we are looking at the limit as    goes to negative infinity, we have  
So, we have

       

Step 2:  
We simplify to get

       


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

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