Limit as x-> 4 of 3-x / x^2-2x-8

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In summary, the limit of (3-x)/(x^2-2x-8) as x approaches 4 from the left is positive infinity, as the numerator approaches -1 and the denominator approaches 0- which results in a fraction of -1/0- that equals infinity. However, the limit does not exist as the two-sided limit is positive and negative infinity, indicating that the function has a vertical asymptote at x=4.
  • #1
mooneh
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lim 3-x / x^2-2x-8
x-> 4-


the answer is + infinit
 
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  • #2
since as X -> 4-
the numerator will go to -1
and the denominator will go to 0- because x is just a little less than 4

the fraction -1/0- (when 0 is on the left side of the real axis and really close to 0) will give you infinity
 
  • #3
mooneh said:
lim 3-x / x^2-2x-8
x-> 4-


the answer is + infinit
Please use parentheses! What you wrote was 3- (x/x^2)- 2x- 8= 3- 1/x- 2x- 8. What you really meant (I think!) was (3- x)/(x^2- 2x- 8). At= x= 4, the numerator is 3- 4= -1 and the denominator is [itex]4^2- 2(4)- 8= 16- 8- 8= 0[/itex]. That much tells you the (two-sided) limit does not exist- the two one-sided limits must be positve and negative infinity. If x< 4, say x= 3, then the denominator is [itex]3^2- 2(3)- 8= 9- 6- 8= - 5[/itex]. The crucial point is that it is negative. Since a polynomial can only change signs at points where its value is 0 ([itex]x^2- 2x- 8= (x+ 2)(x- 4)[/itex] that can only happen at x= -2 and x= 4), the denominator must negative for -2< x< 4 and x. The numerator is close to -1 for any x close to 4 so the fraction is positive for all -2< x< 4. That tells you the limit is ____________.
 

FAQ: Limit as x-> 4 of 3-x / x^2-2x-8

What is the limit as x approaches 4 of the given function?

The limit as x approaches 4 of the function 3-x / x^2-2x-8 can be found by plugging in 4 for x in the function. This results in a limit of -1/4, meaning that as x gets closer and closer to 4, the function approaches a value of -1/4.

How do you solve for the limit as x approaches 4?

To solve for the limit as x approaches 4, you can use direct substitution by plugging in 4 for x in the function. However, if this results in an undefined answer, you may need to use algebraic manipulation or other methods such as factoring or L'Hopital's rule to find the limit.

What does the limit as x approaches 4 tell us about the function?

The limit as x approaches 4 of a function tells us the value that the function approaches as x gets closer and closer to 4. It does not necessarily tell us the actual value of the function at x=4, but it can give us insight into the behavior of the function near that point.

Can the limit as x approaches 4 be different from the actual value of the function at x=4?

Yes, the limit as x approaches 4 and the actual value of the function at x=4 can be different. This is because the limit only tells us the behavior of the function as x gets closer and closer to 4, while the actual value is the specific value of the function at that point.

How can we use the limit as x approaches 4 to determine if the function is continuous?

If the limit as x approaches 4 and the actual value of the function at x=4 are the same, then the function is said to be continuous at x=4. However, if the limit and the actual value differ, the function is not continuous at that point. This means that there is a jump or a hole in the graph of the function at x=4.

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