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# Limits

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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

#2
+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:

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

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Sep 12, 2019

#1
-1

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

Sep 12, 2019
#2
+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:

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

hectictar Sep 12, 2019