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\(\text{Is there a number $a$ such that the limit}\\ lim_{x\to 3}\frac{2x^2-3ax+x-a-1}{x^2-2x-3}\,\text{exists?}\)

 Sep 12, 2019

Best Answer 

 #2
avatar+8720 
+1

We want to find  a  such that the numerator is also zero when  x = 3

 

That is, we want to find  a  such that...

 

2(3)2 - 3a(3) + 3 - a - 1  =  0

 

18 - 9a + 3 - a - 1  =  0

 

20 - 10a  =  0

 

20  =  10a

 

2  =  a

 

So yes, when  a = 2,  the limit exists.

 

Check:

 

\(\lim\limits_{x\to3}\frac{2x^2-6x+x-3}{x^2-2x-3}\ =\ \lim\limits_{x\to3}\frac{2x(x-3)+1(x-3)}{(x-3)(x+1)}\ =\ \lim\limits_{x\to3}\frac{2x+1}{x+1}\ =\frac74\)

.
 
 Sep 12, 2019
 #1
avatar
-1

 

It's obvious that, no matter what the value of "a" is, when x=3 the denominator becomes zero. 

 
 Sep 12, 2019
 #2
avatar+8720 
+1
Best Answer

We want to find  a  such that the numerator is also zero when  x = 3

 

That is, we want to find  a  such that...

 

2(3)2 - 3a(3) + 3 - a - 1  =  0

 

18 - 9a + 3 - a - 1  =  0

 

20 - 10a  =  0

 

20  =  10a

 

2  =  a

 

So yes, when  a = 2,  the limit exists.

 

Check:

 

\(\lim\limits_{x\to3}\frac{2x^2-6x+x-3}{x^2-2x-3}\ =\ \lim\limits_{x\to3}\frac{2x(x-3)+1(x-3)}{(x-3)(x+1)}\ =\ \lim\limits_{x\to3}\frac{2x+1}{x+1}\ =\frac74\)

 
hectictar Sep 12, 2019

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