Finding a limit of an expression when the denominator tends to zero.

[TheSodesa]

Homework Statement

This problem is found in a chapter about rational functions in my review book. The expression is as follows:

<br /> \lim_{x\rightarrow a} {\frac{2 x^2 + 5x - 3}{x - a}} = \ \text{L}<br />

where 'a' is a constant. I'm supposed to find a value of 'a' that allows the limit to exist.

Homework Equations

The only advice given in the book is to try and simplify an expression into a from where it is defined, should the denominator tend to zero.

The Attempt at a Solution

I tried to see if you could simplify the expression into a form where it would be defined, but no such luck. No common factors anywhere I could see.

The solution at the end of the book states, that the expression should be of the form 0/0, or in other words the constant 'a' should be an x intercept of the numerator, and I'm unsure as to why this is the case. Since the book I'm using is a review book, it doesn't contain much in terms of theory, so I'm out of luck in terms of that.

Can anybody spare the time to explain this to me, even though it seems there might be a lot to explain?

 

[thelema418]

Did you try factoring the numerator?

 

[TheSodesa]

Did you try factoring the numerator?

Factoring the numerator yields either:

(2x - 1)(x + 6) \ \text{or} \ (2x + 6)(x - \frac{1}{2})

Inputting a to these two, we get:

(2a - 1)(a + 6) \ \text{or} \ (2a + 6)(a - \frac{1}{2})

This is all fine and dandy, except it hasn't brought me closer to the solution. Why should the numerator be zero in the limit, like the answer insists?

According to the book:a = -3 \ \text{or} \ a = \frac{1}{2}

If a = -3, the limit is -7, and if a = \frac{1}{2}, the limit is 7.

"When x = a, the expression should be of the form \frac{0}{0}, or in other words, 'a' should be one of the x-intercepts of the numerator, -3 \ \text{or} \ \frac{1}{2}"

Why does 'a' have to be one of the numerator's x-intercepts? There is something fundamental about rational limits I'm not getting here.

 

[TheSodesa]

Factoring the numerator yields either:

(2x - 1)(x + 6) \ \text{or} \ (2x + 6)(x - \frac{1}{2})

Inputting a to these two, we get:

(2a - 1)(a + 6) \ \text{or} \ (2a + 6)(a - \frac{1}{2})

This is all fine and dandy, except it hasn't brought me closer to the solution. Why should the numerator be zero in the limit, like the answer insists?

According to the book:a = -3 \ \text{or} \ a = \frac{1}{2}

If a = -3, the limit is -7, and if a = \frac{1}{2}, the limit is 7.

"When x = a, the expression should be of the form \frac{0}{0}, or in other words, 'a' should be one of the x-intercepts of the numerator, -3 \ \text{or} \ \frac{1}{2}"

Why does 'a' have to be one of the numerator's x-intercepts? There is something fundamental about rational limits I'm not getting here.

Looking at the factorized form of the numerator, I now see that the answers given by the book would qualify as the roots of the polynomial in the numerator. That still doesn't tell me why I should form an equation, where the limit is equal to zero.

Why should the limit be equal to zero?

 

[PeroK]

Deleted.

 

[PeroK]

Homework Statement

This problem is found in a chapter about rational functions in my review book. The expression is as follows:

<br /> \lim_{x\rightarrow a} {\frac{2 x^2 + 5x - 3}{x - a}} = \ \text{L}<br />

where 'a' is a constant. I'm supposed to find a value of 'a' that allows the limit to exist.

If a is not a root of the numerator, then the limit of the numerator is not 0, so the limit of the fraction does not exist.

 

[TheSodesa]

If a is not a root of the numerator, then the limit of the numerator is not 0, so the limit of the fraction does not exist.

This is what I'm having trouble grasping. Even if the numerator was 0, the fraction would be in the form \frac{0}{0}, so we would still be dividing by zero, and the limit wouldn't exist.

 

[PeroK]

This is what I'm having trouble grasping. Even if the numerator was 0, the fraction would be in the form \frac{0}{0}, so we would still be dividing by zero, and the limit wouldn't exist.

You are missing the key point about limits of the form 0/0. The limit is for values of x close to a, but not equal to a. So, you are not evaluating 0/0 but a fraction where both numerator and denominator are close to 0.

Consider:

lim_{x → 0} \frac{x}{x} \ or \ lim_{x → 0} \frac{x^2}{x}

These limits are for x getting close to (but not equal to) 0. Both exist.

 

[TheSodesa]

You are missing the key point about limits of the form 0/0. The limit is for values of x close to a, but not equal to a. So, you are not evaluating 0/0 but a fraction where both numerator and denominator are close to 0.

Consider:

lim_{x → 0} \frac{x}{x} \ or \ lim_{x → 0} \frac{x^2}{x}

These limits are for x getting close to (but not equal to) 0. Both exist.

Alright.

Could you then please clarify why a limit, this particular limit, wouldn't exist if the limit of the numerator was not zero?

Why is the form \frac{0}{0} of a fraction so important, and why does it allow us to compute limits by substitution?

 

[TheSodesa]

You are missing the key point about limits of the form 0/0. The limit is for values of x close to a, but not equal to a. So, you are not evaluating 0/0 but a fraction where both numerator and denominator are close to 0.

Consider:

lim_{x → 0} \frac{x}{x} \ or \ lim_{x → 0} \frac{x^2}{x}

These limits are for x getting close to (but not equal to) 0. Both exist.

An additional question: Why do you say "the limit of the numerator"? Aren't we concerned about the whole limit, including the denominator?

 

[PeroK]

Alright.

Could you then please clarify why a limit, this particular limit, wouldn't exist if the limit of the numerator was not zero?

You're starting to confuse yourself now. If the numerator does not tend to 0 but the denominator does tend to 0, then clearly the limit will diverge to +/-∞.

Why is the form \frac{0}{0} of a fraction so important, and why does it allow us to compute limits by substitution?

Any other limit allows us simply to substitute (assuming both numerator and denominator are continuous functions). 0/0 is the tricky one, because the limit may exist or may not exist. In this case, you have to calculate what is happening close to, but not equal to, the limit point.

 

[TheSodesa]

You're starting to confuse yourself now. If the numerator does not tend to 0 but the denominator does tend to 0, then clearly the limit will diverge to +/-∞.

Ahh, of course. How did I not see that?

http://www.wolframalpha.com/input/?i=graph+lim+1/x+as+x+approaches+a

http://www.wolframalpha.com/input/?i=graph+lim+-1/x+as+x+approaches+a

=> If the denominator tends to zero, the numerator also has to tend to zero, or else the limit will grow without limit, pun intended , and we wouldn't be able to calculate a 2-sided limit.

Any other limit allows us simply to substitute (assuming both numerator and denominator are continuous functions). 0/0 is the tricky one, because the limit may exist or may not exist. In this case, you have to calculate what is happening close to, but not equal to, the limit point.

So after realizing that the numerator also has to tend to zero, it would be simple thing to substitute a for x, set the numerator to equal zero, and see what values of 'a' would satisfy that condition. I'm starting to see what I need to do and why.

I will report back later if I've solved the problem. I need to go now.

 

[pbuk]

I think this has got a little off track. Can I go back to the beginning?

When you have a term that tends towards zero in the denominator the aim is to get the same term in the numerator. Can you see that $$ \lim_{x \rightarrow a} \frac{(x - a)f(x)}{(x - a)} $$ can easily be found?

Now any quadratic expression $A x^2 + B x + C$ can be rewritten as $k(x - a)(x - b)$ and you have tried to do this, although your first solution gives $(2x - 1)(x + 6) = 2x^2 + 11x - 6$ which is not what we want. Instead we have

$$ \lim_{x \rightarrow a} \frac{2x^2 + 5x - 3}{(x - a)} = \lim_{x \rightarrow a} \frac{2(x - (-3))(x - \frac12)}{(x - a)} $$

Can you see how this works now?

 

[PeroK]

Now any quadratic expression $A x^2 + B x + C$ can be rewritten as $k(x - a)(x - b)$

k = A

And, not every quadratic can be factorised with real coefficients.

 

[HakimPhilo]

Write the limit as $\lim\limits_{x\to a}\dfrac{2x^2+5x-3}{x-a}=\lim\limits_{x\to a}{(2x^2+5x-3)}\dfrac1{x-a}$ and consider the limit as $x\to a^+$ and as $x\to a^-$. What do you see?

 

[thelema418]

Factoring the numerator yields either:

(2x - 1)(x + 6) \ \text{or} \ (2x + 6)(x - \frac{1}{2})
.

Well, that's where your problem lies. That factorization is not correct. Notice:

$(2x -1)(x+6) = 2x^2 - x +12x -6 \neq 2x +5x - 3$

Try again or just use the quadratic formula to find the roots. =)

 

[verty]

So after realizing that the numerator also has to tend to zero, it would be simple thing to substitute a for x, set the numerator to equal zero, and see what values of 'a' would satisfy that condition. I'm starting to see what I need to do and why.

I will report back later if I've solved the problem. I need to go now.

Another way of thinking about it is, you want to cancel out the denominator. The problem is dividing by zero, so if you cancel out the denominator, you can take the limit. This is just another way to think about it.

 

[SammyS]

Factoring the numerator yields either:

(2x - 1)(x + 6) \ \text{or} \ (2x + 6)(x - \frac{1}{2})

Inputting a to these two, we get:

(2a - 1)(a + 6) \ \text{or} \ (2a + 6)(a - \frac{1}{2})

This is all fine and dandy, except it hasn't brought me closer to the solution. Why should the numerator be zero in the limit, like the answer insists?
...

Recall that the solution in the book states that "the expression should be of the form 0/0".

So solve (2a - 1)(a + 6) =0 for a.

Of course, there are two solutions for this.

 

[thelema418]

Recall that the solution in the book states that "the expression should be of the form 0/0".

So solve (2a - 1)(a + 6) =0 for a.

Of course, there are two solutions for this.

Please not my previous post. The OP's issue is with the factoring. The above information is not a correct factoring.

 

[SammyS]

Please note my previous post. The OP's issue is with the factoring. The above information is not a correct factoring.

@thelema418,

Sorry, I missed that.

Of course you're right!

$(2a - 1)(a + 6) =2a^2+11a-6$ not $2a^2+5a-3\ .$
​

But the other factorization looks good:

$\displaystyle (2a + 6)(a - \frac{1}{2})=2a^2+5a-3\ .$

Can OP solve $\displaystyle (2a + 6)(a - \frac{1}{2})=0\ ?$

 

[TheSodesa]

Well, that's where your problem lies. That factorization is not correct. Notice:

$(2x -1)(x+6) = 2x^2 - x +12x -6 \neq 2x +5x - 3$

Try again or just use the quadratic formula to find the roots. =)

Yeah, I totally messed up with the factorization, because I didn't take out the common factor 2, and forgot it is multiplied into BOTH terms. That's what you get when you don't sleep enough.

However, my problem wasn't really with the factorization, since I do have a basic understanding of the theory behind it. What I didn't understand was how to deal with limits, and why I'm allowed to set the numerator to 0, or more importantly, why I'm allowed to just say that the numerator's value should change independently from the denominator's value, as x varies, since they both are dependent on x.

Anyways, I have now solved for a, and got the values stated in the solution (-3 and 1/2).

Plugging those values in, we get:

\frac{2(x+3)(x-\frac{1}{2})}{x+3} = 2(x-\frac{1}{2})\text{ as a = -3}

and

\frac{2(x+3)(x-\frac{1}{2})}{x-\frac{1}{2}} = 2(x+3)\text{ as a = 1/2}

In both cases, we get rid of the denominator, and the divide-by-zero problem disappears.

<br /> \lim_{x\rightarrow -3} {\frac{2 x^2 + 5x - 3}{x + 3}} = 2(-3-\frac{1}{2}) = -7<br />

<br /> \lim_{x\rightarrow \frac{1}{2}} {\frac{2 x^2 + 5x - 3}{x - \frac{1}{2}}} = 2(\frac{1}{2}+3) = 7<br />

Boom, solved.

 

[HallsofIvy]

Generally, to find \lim_{x\to a} f(x)/g(x) where f and g are continuous at a (especially if f and g are polynomials) start by finding \lim_{x\to a} f(x) and \lim_{x\to a} g(x).

1) if \lim_{x\to a} g(x)= g(a)\ne 0 then the limit is f(a)/g(a).

2) if \lim_{x\to a} g(x)= g(a)= 0 and \lim_{x\to a} f(x)= f(a)\ne 0 then the limit does not exist.

3) if \lim_{x\to a} g(x)= g(a)= 0 and \lim_{x\to a} f(x)= f(a)= 0 then we have to look more closely. If f(x) and g(x) are polynomials, the fact that f(a)= g(a)= 0 tells us that we can factor x- a out of both.

 

[thelema418]

What I didn't understand was how to deal with limits, and why I'm allowed to set the numerator to 0, or more importantly, why I'm allowed to just say that the numerator's value should change independently from the denominator's value, as x varies, since they both are dependent on x.

Anyways, I have now solved for a, and got the values stated in the solution (-3 and 1/2).

Plugging those values in, we get:

\frac{2(x+3)(x-\frac{1}{2})}{x+3} = 2(x-\frac{1}{2})\text{ as a = -3}

and

\frac{2(x+3)(x-\frac{1}{2})}{x-\frac{1}{2}} = 2(x+3)\text{ as a = 1/2}

In both cases, we get rid of the denominator, and the divide-by-zero problem disappears.

<br /> \lim_{x\rightarrow -3} {\frac{2 x^2 + 5x - 3}{x + 3}} = 2(-3-\frac{1}{2}) = -7<br />

<br /> \lim_{x\rightarrow \frac{1}{2}} {\frac{2 x^2 + 5x - 3}{x - \frac{1}{2}}} = 2(\frac{1}{2}+3) = 7<br />

Boom, solved.

Your problem creates a situation where you must divide by 0. Note that

1) A non-zero number divided by 0 is undefined.
2) 0 divided by 0 is indeterminate.

There are an infinite number of situations where you are dividing by 0 in this problem. When we find the roots of the numerator, we are isolating the situation that gives 0/0 because this is the only place where it is possible to have a limit.

I think you need to look at the limit from the left and the right.

Consider:
$\lim_{x\rightarrow -3^-} {\frac{(2x-1)(x+3)}{(x + 3)}}$

We are approaching from the left, so x is a number less the -3.
$x < -3$
$x + 3< 0$
So we know $x + 3$ is a positive integer k. Now we have
$\lim_{x\rightarrow -3^-} {\frac{(2x-1)k}{k}} = (2x - 1)$

Now you have to look at $-3^+$. You want to show that this limit is the same as the previous limit.

 

[TheSodesa]

Your problem creates a situation where you must divide by 0. Note that

1) A non-zero number divided by 0 is undefined.
2) 0 divided by 0 is indeterminate.

There are an infinite number of situations where you are dividing by 0 in this problem. When we find the roots of the numerator, we are isolating the situation that gives 0/0 because this is the only place where it is possible to have a limit.

I think you need to look at the limit from the left and the right.

Consider:
$\lim_{x\rightarrow -3^-} {\frac{(2x-1)(x+3)}{(x + 3)}}$

We are approaching from the left, so x is a number less the -3.
$x < -3$
$x + 3< 0$
So we know $x + 3$ is a positive integer k. Now we have
$\lim_{x\rightarrow -3^-} {\frac{(2x-1)k}{k}} = (2x - 1)$

Now you have to look at $-3^+$. You want to show that this limit is the same as the previous limit.

Hmm, I suppose this is a good way of looking at it if you want to be extremely thorough. For my purposes, however, this is a bit too detailed of an answer.

Thanks for the response anyways.