Find Limit of (x^2+1)/(x^2-4x+3) as x->3

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In summary, the limit of (x^2+1)/(x^2-4x+3) as x approaches 3 does not exist. This can be verified by putting the equation into a calculator or by factoring the denominator and observing the behavior of the function as x approaches 3 from either side. The function is not continuous at this point, as the one-sided limits approach positive and negative infinity. It is important to first test if the numerator reduces to 0/0, but in this case, it does not, indicating a vertical asymptote.
  • #1
Seldini
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lim x->3

(x^2+1)/(x^2-4x+3)

Can anyone show me how to find the limit of this one? Thanks guys.
 
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  • #2
Seldini said:
lim x->3

(x^2+1)/(x^2-4x+3)

Can anyone show me how to find the limit of this one? Thanks guys.

is this from a certain side? (ie 3+ or 3- ?)

do you know what this graph looks like? is it a continuous function?
 
  • #3
For a rational function of the form

[tex]\frac{f(x)}{g(x)}[/tex]

provided [tex]g(c)\neq 0[/tex] and provided f(x) and g(x) are polynomials as in your case you can write

[tex]Lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \frac{Lim_{x \rightarrow c}f(x)}{Lim_{x \rightarrow c}g(x)}=\frac{f(c)}{g(c)}[/tex]

Cheers
vivek
 
Last edited:
  • #4
g(c)=0 in this case

I recommend you just graph it or make a table of values
 
  • #5
The limit does not exist. There are some ways you can look at this. First of all, you can just put it into a calculator and you'll see it does not exist.

Another solution is,
the number will always be positive. The denominator however will not be. You can factor it and it becomes (x - 1)(x - 3). If x is a little bit less then 3 then the denominator is negative. If x is a little greater then 3 the denominator is positive. Hence, if youre approaching from the left side, the one-sided limit is negative infinite. From the right, it is positive infinite. As you get closer and closer to 3 from either side the magnitude constant increases because the denominator gets closer and closer to zero. Hence, on either side it will constantly increase but in different directions.

Is that valid? Or is that just verbal diarrhea?
 
  • #6
Parth Dave said:
The limit does not exist. There are some ways you can look at this. First of all, you can just put it into a calculator and you'll see it does not exist.

Another solution is,
the number will always be positive. The denominator however will not be. You can factor it and it becomes (x - 1)(x - 3). If x is a little bit less then 3 then the denominator is negative. If x is a little greater then 3 the denominator is positive. Hence, if youre approaching from the left side, the one-sided limit is negative infinite. From the right, it is positive infinite. As you get closer and closer to 3 from either side the magnitude constant increases because the denominator gets closer and closer to zero. Hence, on either side it will constantly increase but in different directions.

Is that valid? Or is that just verbal diarrhea?

yes, the limit does not exist. the function is not continuous at that point, ie the limit from one side approaches negative infinity, the limit from the other side approaches infinity.
 
  • #7
Parth Dave said:
The limit does not exist. There are some ways you can look at this. First of all, you can just put it into a calculator and you'll see it does not exist.

Another solution is,
the number will always be positive. The denominator however will not be. You can factor it and it becomes (x - 1)(x - 3). If x is a little bit less then 3 then the denominator is negative. If x is a little greater then 3 the denominator is positive. Hence, if youre approaching from the left side, the one-sided limit is negative infinite. From the right, it is positive infinite. As you get closer and closer to 3 from either side the magnitude constant increases because the denominator gets closer and closer to zero. Hence, on either side it will constantly increase but in different directions.

Is that valid? Or is that just verbal diarrhea?

Yes, it is valid. The only thing you left out was to first test and see if f(c) reduces to 0/0. If it does, you still have a limit, even though the function's not continuous. In this case, the numerator is equal to 10 at f(3), so you know you have a vertical assymptote and Parth Dave's analysis is right on the money.

A whole lot of common sense beats algebra on most days.
 

1. What is the limit of (x^2+1)/(x^2-4x+3) as x approaches 3?

The limit of (x^2+1)/(x^2-4x+3) as x approaches 3 is undefined. This is because when x=3, the denominator becomes 0, which is undefined in mathematics.

2. How do you find the limit of (x^2+1)/(x^2-4x+3) as x approaches 3?

To find the limit of (x^2+1)/(x^2-4x+3) as x approaches 3, we can use the direct substitution method. This means plugging in x=3 into the function and simplifying. However, since this results in an undefined value, we can also use algebraic manipulation or graphing to determine the limit.

3. Can you use L'Hopital's rule to find the limit of (x^2+1)/(x^2-4x+3) as x approaches 3?

Yes, L'Hopital's rule can be used to find the limit of (x^2+1)/(x^2-4x+3) as x approaches 3. This rule states that if the limit of the quotient of two functions is in an indeterminate form (such as 0/0 or ∞/∞), then the limit of the original function can be found by taking the derivative of the numerator and denominator and evaluating the new quotient at the given value.

4. Is there a way to simplify the expression (x^2+1)/(x^2-4x+3) before finding the limit as x approaches 3?

Yes, the expression (x^2+1)/(x^2-4x+3) can be simplified by factoring the denominator. This results in (x^2+1)/(x-3)(x-1). However, this does not change the limit as x approaches 3. It only makes it easier to evaluate the function at that specific point.

5. What is the significance of finding the limit of (x^2+1)/(x^2-4x+3) as x approaches 3?

Finding the limit of (x^2+1)/(x^2-4x+3) as x approaches 3 allows us to understand the behavior of the function near the point x=3. It tells us what value the function is approaching as x gets closer and closer to 3. This information can be useful in many mathematical and scientific applications, such as in optimization problems or in understanding the behavior of a physical system.

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