# Homework Help: Finding a limit of an expression when the denominator tends to zero.

1. Jul 18, 2014

### TheSodesa

1. The problem statement, all variables and given/known data

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

$$\lim_{x\rightarrow a} {\frac{2 x^2 + 5x - 3}{x - a}} = \ \text{L}$$

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

2. Relevant 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.

3. 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?

Last edited: Jul 18, 2014
2. Jul 18, 2014

### thelema418

Did you try factoring the numerator?

3. Jul 18, 2014

### 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.

Last edited: Jul 18, 2014
4. Jul 18, 2014

### TheSodesa

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?

Last edited: Jul 18, 2014
5. Jul 18, 2014

### PeroK

Deleted.

6. Jul 18, 2014

### PeroK

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.

7. Jul 18, 2014

### TheSodesa

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.

8. Jul 18, 2014

### PeroK

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.

9. Jul 18, 2014

### TheSodesa

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?

10. Jul 18, 2014

### TheSodesa

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

11. Jul 18, 2014

### PeroK

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 +/-∞.

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.

12. Jul 18, 2014

### TheSodesa

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 :tongue:, and we wouldn't be able to calculate a 2-sided limit.

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.

13. Jul 18, 2014

### MrAnchovy

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?

Last edited: Jul 18, 2014
14. Jul 18, 2014

### PeroK

k = A

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

15. Jul 18, 2014

### 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?

Last edited: Jul 18, 2014
16. Jul 18, 2014

### thelema418

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. =)

17. Jul 18, 2014

### verty

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.

18. Jul 18, 2014

### SammyS

Staff Emeritus
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.

19. Jul 18, 2014

### thelema418

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

20. Jul 18, 2014

### SammyS

Staff Emeritus
@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 $\displaystyle (2a + 6)(a - \frac{1}{2})=0\ ?$