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I am taking calculus again and I am trying to evaluate limits.

Anyways here is the first one I had trouble with $$\lim_{x\to 5}\frac{x^2-5x+6}{x-5}.$$ I know that the answer is "Does not exist" but I don't know how to determine that.

Next is something I have no idea how to deal with, I tried quadratic formula (which is of course useless which I now know), conjugates, factoring down further but I can't seem to manipulate the denominator in any way that is useful. $$\lim_{x\to -3}\frac{t^2-9}{2t^2+7t+3}.$$

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No, these are all suppose to be done algebraically. – user138246 Jan 14 '12 at 22:19
Are you supposed to use epsilon-delta proofs or general "non-rigorous" reasoning? In either case, hint for the second one: the denominator factors as (t+3)(2t+1). – Lopsy Jan 14 '12 at 22:24
Can you see $\dfrac{x^2-5x+6}{x-5} = x + \dfrac{6}{x-5}$ and hence the problem as ${x\to 5}$? – Henry Jan 14 '12 at 22:33
Yes I can see that but if I were to see that before I still would have just tried manipulating the problem in a way to make the 0 on the bottom to dissapear. – user138246 Jan 14 '12 at 22:34
up vote 3 down vote accepted

To factor the numerator of the first limit, all you have to do is find the roots of the quadratic (use the quadratic formula, if everything else fails). Here, the two roots are $2$ and $3$, so $x^2-5x+6 = (x-2)(x-3)$.

But the key point is to think about what happens as $x$ approaches $5$. As $x$ approaches $5$, the denominator is getting smaller, and smaller, and smaller, approaching $0$. On the other hand, the numerator approaches $(5)^2 -5(5) + 6 = 6$. If you have a number that is getting closer and closer and closer to $6$, divided by a number that is getting closer and closer and closer to $0$, you are going to get numbers that are larger, and larger, and larger, and larger in size (big and positive if $x$ is bigger than, but very close to, $5$, and big and negative if $x$ is less than, but very close to, $5$). That's why the limit does not exist. Nothing to do with factoring.

This kind of situation will hold whenever you have a quotient, the denominator approaches zero, and the numerator does not approach zero. Because the values of the quotient will be getting larger and larger and larger in size as you approach the limit.

The second limit is more interesting: here, both the numerator and the denominator approach $0$, so the line of argument used above does not work. Here, it is possible for the limit to exist. And because what you have are polynomials, now is the time to do some factoring.

$t^2-9$ is a difference of squares, so it factors as $(t-3)(t+3) = (t-3)(t-(-3))$. That second factor, $(t-(-3))$, is the reason why the numerator approaches $0$ as $t$ approaches $-3$.

The denominator also approaches $0$ as $t$ approaches $-3$, so again it must be that there is a factor of $(t-(-3))$ "hiding". You can figure out how to factor using long division, or by finding factoring out the leading coefficient and using the quadratic formula. According to the quadratic formula, the roots of $2t^2 + 7t + 3$ are $$\frac{-7 \pm\sqrt{7^2 - 4(2)(3)}}{2(2)} = \frac{-7\pm\sqrt{49-24}}{4} = \frac{-7\pm\sqrt{25}}{4} = \frac{-7\pm 5}{4} = \left\{\begin{array}{r} -\frac{1}{2}\\ -3 \end{array}\right.$$ So that means that $$\begin{align*} 2t^2 +7t + 3 &= 2\left(t^2 + \frac{7}{2}t + \frac{3}{2}\right)\\ &= 2\left(t -\left(-\frac{1}{2}\right)\right)\left(t - (-3)\right)\\ &= 2\left(t+\frac{1}{2}\right)\left(t+3\right)\\ &= \left(2t + 1\right)(t+3). \end{align*}$$

So now you have $$\lim_{t\to-3}\frac{t^2-9}{2t^2 +7t + 3} = \lim_{t\to-3}\frac{(t-3)(t+3)}{(2t+1)(t+3)}.$$

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I don't quite follow the last part after "So that means that" what is happening? – user138246 Jan 14 '12 at 22:27
@Jordan: If $t^2+at+b$ is a quadratic polynomial, and $r_1$ and $r_2$ are its two roots, then $t^2+at+b = (t-r_1)(t-r_2)$. So first I factored out $2$ to get a polynomial of that form (with $a=\frac{7}{2}$, $b=\frac{3}{2}$); then I used the fact that the two roots were $r_1=-\frac{1}{2}$ and $r_2=-3$; and then I distributed the $2$ into the first parenthesis to get rid of the fraction. Pure algebra, no calculus anywhere. – Arturo Magidin Jan 14 '12 at 22:33
Thanks I see it now. Calculus isn't my problem it all seems very easy to me my problem is algebra. I can never remember all the little tricks and rules. I know these are all things I am supposed to understand but no matter what I do I never come to that understanding where I just remember these things forever, I always forget. – user138246 Jan 14 '12 at 22:35

For what it's worth:

I'll just address how to factor quadratics.

First, you need to master performing multiplications of the form $(ax+b)(cx+d)$. Factoring is essentially doing this in reverse.

As for factoring:

Here's one way, by example, to factor certain quadratics; namely, where the quadratic factors as $(ax+b)(cx+d)$, where $a,b,c,d$ are all integers.

To factor $$2t^2+7t+3,$$ I would obtain the factorization by just "guessing and checking":

You first write $$\tag{1} 2t^2+7t+3=(at+b)(ct+d). $$

You first guess values for $a$ and $c$. This is an educated guess: The product $ac$ has to be 2 because if you multiplied the right hand side of (1) out, the $t^2$ term will be $ac\cdot t^2$ and the $t^2$ term on the left is just $2t ^2$.

Here's the guess: $a=2, c=1$. Let's try that. This gives us

$$\tag{2} 2t^2+7t+3=(2t+b)(t+d). $$

Next, we guess the values of $b$ and $d$. But, it's an educated guess. Because the product $bd$ has to be 3 (the constant term on the left hand side of $(2)$ is 3), we guess

$$\tag{3} b= 3, d= 1, \quad \text{or} \quad b=-3, d=-1 $$ (note we made two guesses at the same time. We know since the product is positive 3, that $b$ and $d$ have the same signs. So we guess magnitudes and try all possible sign combinations).

Then we have

$$\tag{4} 2t^2+7t+3=(\color{darkgreen}{2t}+\color{maroon}{\pm 3})(\color{maroon}{t}+\color{darkgreen}{\pm1}). $$

Here comes the "check" part: The $t$ term on both sides of $(4)$, after expansion, has to be $7t$. This is obvious looking at the left hand side. How is the $t$ term on the right hand side obtained? Well from two parts: it's the sum of the product of the maroon quantities and the product of the green quantities. You need to decide if any choices of sign will give you $7t$. Here we have, taking the products and summing: $$ \color{darkgreen}{2t}\cdot (\color{darkgreen}{+1})+(\color{maroon}{+3})\cdot \color{maroon}t=5t ,\qquad \color{darkgreen}{2t}\cdot (\color{darkgreen}{-1})+(\color{maroon}{-3})\cdot \color{maroon}t=-5t . $$ None work...

But that's not altogether bad, at least we know the guesses in (3) do not work.

Let's try a different guess:

$$ 2t^2+7t+3=(\color{darkgreen}{2t}+\color{maroon}{\pm 1})(\color{maroon}{t}+\color{darkgreen}{\pm3}). $$

Here, if we chose $ \color{maroon}{+1}$ and $\color{darkgreen}{+3}$:

$$ 2t^2+7t+3=(\color{darkgreen}{2t}+\color{maroon}{ 1})(\color{maroon}{t}+\color{darkgreen}{ 3}) $$ works (check)! And we are done.

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