# Thread: Finding lim(x->2) of following expression

1. ## Finding lim(x->2) of following expression

I have encountered a problem I have been given that I cannot solve. The problem is:

lim(x->2) {1/(x-2) - 1/(x^2-4) }

After a good couple of hours of trying to manipulate the expression to see if I could solve it, I got frustrated and graphed the function, and realized that the limit does not exist.

To save such frustration in the future, is there any way of determining from an expression if the limit actually exists, before trying to solve it?

2. Originally Posted by borophyll
To save such frustration in the future, is there any way of determining from an expression if the limit actually exists, before trying to solve it?
There is no really good answer to your question.
But I would have noticed that $\frac{1}{{x - 2}} - \frac{1}{{x^2 - 4}} = \frac{{x + 1}}{{x^2 - 4}}$.
The more of these you do will help you.

3. Easily enough, if you let $u= x - 2$ and $v = x+2$ then

$\lim_{x\to2} \frac{1}{x-2} - \frac{1}{(x-2)(x+2)} = \lim_{u\to0} \frac{1}{u} - \frac{1}{uv} = \lim_{u\to0} \frac{v-1}{uv} = \lim_{u\to0} \frac{v-1}{v} * \frac{1}{u} = \infty$

Since $v,v-1$ are bounded and $u \rightarrow 0 \Rightarrow \frac{1}{u} \rightarrow \infty$

Generally, if you want to see this, simply look at the denominator -- you can see that as x approaches 2, the denominator approaches 0, so the fraction approaches infinity.

4. find $lim_{x \rightarrow 2^+}$ and find $lim_{x \rightarrow 2^-}$ and see if they are different?

5. you can use L'hopital (if i spelled it correctly)
$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$
if $\lim_{x\rightarrow a},f(x),g(x)=0,\pm\infty$ to the same value, and $f(x),g(x)\in C^1$ in the vecinity of $a$(not the complex numbers, the group of continuse functions that have a continuse first dirivative at the vecinity of a )
then $\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$

so in your problem
$\lim_{x\rightarrow 2}\frac{x+1}{x^2-4}=\lim_{x\rightarrow 2}\frac{1}{2x}\rightarrow \frac{1}{4}$

6. Originally Posted by elfsong
you can use L'hopital (if i spelled it correctly)
$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$
if $\lim_{x\rightarrow a},f(x),g(x)=0,\pm\infty$ to the same value, and $f(x),g(x)\in C^1$ in the vecinity of $a$(not the complex numbers, the group of continuse functions that have a continuse first dirivative at the vecinity of a )
then $\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$
so in your problem
$\lim_{x\rightarrow 2}\frac{x+1}{x^2-4}=\lim_{x\rightarrow 2}\frac{1}{2x}\rightarrow \frac{1}{4}$
Not only did you misspell l’Hospital but you wrongly applied the rule.
That is not an indeterminate form.

7. Originally Posted by Haytham
find $lim_{x \rightarrow 2^+}$ and find $lim_{x \rightarrow 2^-}$ and see if they are different?
In this case, however, the one-sided limits are equal!

elfsong, in this case, $\lim_{x\to2} f(x) = \lim_{x\to2} x+1 = 3$ , so your application of L'Hospital is obviously incorrect!

8. actually I wasnt saying they were different, I was asking
It makes no difference generally, its just simple to see in this case

I only mentioned because it was what he did when he plotted the function

one last thing, they are different

9. I just don't get how you can 'see' from the expression whether it can be solved or not. For example, take the following:

lim(x->-1) {(x^3+1)/(x+1)}

So logically I know you need to get rid of x+1 from the denominator, and the only way to do that is multiply by one (y/y). So I need to chose some value y such that when multiplied by x^3+1, it will generate a polynomial with a factor of x+1. (and obviously the value y cannot be x+1 itself) I have no idea how to figure this out? I think I lack the intuition of being able to see patterns like this. Is there some algorithmic process I can apply to limit problems?

10. Since (x+1) is a factor of x^3+1 (because -1^3+1=0) you know you can factorise it as (x+1)(P(x)) where
P is another polynomial in x. Since you need x^3 when you multiply it out you have to start with x^2. Now you have x^3 +x^2. So then you need -x to get rid of the unwanted x^2. And the last term must be +1. And so x^3+1 = (x+1)(x^2-x+1). You can do any other polynomial you already know one factor for in essentially the same way, and you could certainly formalise this into an algorithm that a computer could implement. There are rules you can learn which tell you how many real roots a polynomial could have, and what integer factors are possible, though personally I can't remember them (except for obvious things like an integer root must be a factor of the constant term) and I would probably sketch a graph of the polynomial and try to work out where the turning points were so that I would know roughly where to look for a zero.
Don't beat yourself up about not having intuition. Intuition develops with practice.