# Thread: Understanding Limits

1. ## Understanding Limits

I'm on my second week of Calculus, and I still cannot seem to grasp what a limit is and what exactly it does. I've tried getting help, but most tutors don't seem to understand it, and the ones who do are almost never available.

Why are they structured as

lim (3x+2)
x --> 2

and

lim tan ((piX)/4)
x --> 3

why are do the x --> c sometimes have the c's: c- or c+?

Basically, what is the point and why are they structured the way they are. I know this is a lot to ask all at once, but I feel I have exhausted all the rather poor resources I have available.

2. Very quickly ... Say you have an expression like $\displaystyle f(x) = \frac {x^2-4}{x-2}$ and you want to find f(2).

Now it's clear that plugging 2 into the expression won't help because you get 0/0 which is undefined.

But if you creep up on it from above, setting x = 3, x = 2.5, x = 2.1, x = 2.01, x = 2.001 ... and so on, you get a sort of picture of what you think the answer is also trying to creep up on, which (as you'll find out when you try it) is 4. That's approaching the limit from above, and can be written:
$\displaystyle \lim_{x \to 2^+} f(x) = 4$

You can do the same thing by creeping up on it from below, setting, for example, x=1, x=1.5, x=1.9, x=1.99, etc., which can be written:

$\displaystyle \lim_{x \to 2^-} f(x) = 4$ (as you'll find out when you try it out).

Now, if it just so happens that your expression works out as the the same whether you creep up on it from above OR below, then you don't need to write which direction you came to it from, and can write:

$\displaystyle \lim_{x \to 2} f(x) = 4$

Okay, so how do we "know" it's 4? Because

$\displaystyle \frac {x^2 - 4} {x-2} = \frac {(x+2)(x-2)} {x-2} = x+2$ but this ONLY HOLDS when $\displaystyle x \ne 2$ because although $\displaystyle \frac {x^2 - 4} {x-2} = x+2$ most of the time, whatever algebraic manipulations you care to play with, you can NOT get $\displaystyle \frac {x^2 - 4} {x-2}$ to have a value when x = 2, ALL YOU CAN DO is say that it "f(x) tends to a limit of 4 when x tends towards 2."

Does this help?

3. Originally Posted by BRen8

Why are they structured as

lim (3x+2)
x --> 2
This structure is here to help you understand that the question is asking what value does the function approach as x approaches 2. It's not asking for a simple substitution although that is how you would solve this problem.

It is solved simply as follows

$\displaystyle \lim_{x \to 2} (3x+2) = (3\times 2+2)= 6+2 = 8$

This must seem very trivial but I would suggest not getting yourself in a twist about it. The idea of a limit as weird as it seems is a nessacary seed that needs to be planted when you move on to finding derivatives via first principles.

Originally Posted by BRen8

why are do the x --> c sometimes have the c's: c- or c+?
This notation indicates what side of the value to take the limit from, the negative sign means a left hand limit and the postitve sign means a right hand limit.

This can be particulary handy when you are finidng a limit in a hybrid function.

4. Ok... So how would you solve a problem like:

lim tan (pi/4)
x-->4

or

lim f(x), where f(x) = { (x^3)+1, x (lesser then) 1
lim-->1 { x+1, x (greater then or equal to) 1

Sorry, don't know how to use the math tags. the "{" wrap around the two equations in the second problem.

5. Originally Posted by BRen8
Ok... So how would you solve a problem like:

lim tan (pi/4)
x-->4
$\displaystyle \lim_{x \to 4} \tan\frac{\pi}{4} = \lim_{ x\to 4} 1 = 1$

This because the function was not a function of x.

Originally Posted by BRen8
lim f(x), where f(x) = { (x^3)+1, x (lesser then) 1
lim-->1 { x+1, x (greater then or equal to) 1

Sorry, don't know how to use the math tags. the "{" wrap around the two equations in the second problem.
In this question if have to determine which function either $\displaystyle x^3+1~,~x<1$ or $\displaystyle x+1~,~x\geq1$ you must substitute 1 into.

The answer is

$\displaystyle \lim_{x \to 1} x+1 = 1+ 1 = 2$

as $\displaystyle x+1~,~x\geq1$ includes 1, as the restriction tells you it is greater and equal to 1.

$\displaystyle x^3+1~,~x<1$ This restriction only wants numbers less than 1. 1 is not less than 1.

6. Originally Posted by pickslides
$\displaystyle \lim_{x \to 4} \tan\frac{\pi}{4} = \lim_{ x\to 4} 1 = 1$

This because the function was not a function of x.

In this question if have to determine which function either $\displaystyle x^3+1~,~x<1$ or $\displaystyle x+1~,~x\geq1$ you must substitute 1 into.

The answer is

$\displaystyle \lim_{x \to 1} x+1 = 1+ 1 = 2$

as $\displaystyle x+1~,~x\geq1$ includes 1, as the restriction tells you it is greater and equal to 1.

$\displaystyle x^3+1~,~x<1$ This restriction only wants numbers less than 1. 1 is not less than 1.
Why is tan (pi/4) = 1?

And

ahh... So its the output of the equation that matters? Not the Input?

7. Originally Posted by BRen8
Why is tan (pi/4) = 1?
Because the angle here is measured in radians. $\displaystyle 2 \pi$ radians is 360 degrees, so $\displaystyle \pi / 4$ radians is 45 degrees.

Sin 45 degrees = cos 45 degrees, so as $\displaystyle \tan \theta = \frac {\sin \theta} {\cos \theta}$ it follows that tan 45 degrees = 1.

So $\displaystyle tan (pi/4) = 1$.

8. Ok... So how would you find a limit graphically? I understand that it has something to do with the line approaching but never touching a point.

9. Originally Posted by BRen8
I understand that it has something to do with the line approaching but never touching a point.
This is a great way to think about a limit. Consider the function $\displaystyle y = \frac{1}{x}$

When $\displaystyle x \to \infty$ then $\displaystyle y \to 0$ . It does not ever reach zero but it "approaches" it.