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

[Seldini]

lim x->3

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

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

 

[teclo]

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?

 

[maverick280857]

For a rational function of the form

\frac{f(x)}{g(x)}

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

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)}

Cheers
vivek

 

[kreil]

g(c)=0 in this case

I recommend you just graph it or make a table of values

 

[Parth Dave]

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?

 

[teclo]

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.

 

[BobG]

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.