# Calculus, limits?

So far in lecture, we have went over limits, infinite limits, and limits at infinity. My question is how do you know when your answer is suppose to be a negative or positive infinitive, or a integer? I understand one sided limits (thanks to youtube) and that if the left and right limits are different, the limit does not exist. My professor makes the lecture so confusing, and even when I ask questions, he still dont make it clear enough. All his tests are short answer, so if someone can kindly clear this up for me.

### 3 Answers

- MathmomLv 73 years agoFavorite Answer
If you get 0 in numerator and a non-zero in denominator, limit = 0

If you get a non-zero in numerator and 0 in denominator, limit = ±∞ (I'll show how to determine which below)

If you get a non-zero in both numerator and denominator, then limit will be a non-zero number.

If you get 0 in both numerator and denominator, there's no way to determine by that result alone what limit should be. But sometimes you can simplify, so that you get one of the results above.

The simplest way is to determine if an infinite limit is positive or negative is to use + or − sign for all 0's

For example:

As x→2− (2 from the left), then (x−2)→0 from negative side:

x ............. x−2

1.9 ......... −0.1

1.99 ....... −0.01

1.999 ..... −0.001

As x→2+ (2 from the right), then (x−2)→0 from positive side:

x ............ x−2

2.1 ......... 0.1

2.01 ....... 0.01

2.001 ..... 0.001

So I would express this as:

lim[x→2−] (x−2) = −0

lim[x→2+] (x−2) = +0

Now this is not much of a problem at this point.

But when you have (x−2) in denominator, it becomes important:

lim[x→2−] 1/((x−2)(x−5)) = 1/((−0)(−3)) = +∞

lim[x→2+] 1/((x−2)(x−5)) = 1/((+0)(−3)) = −∞

Similarly:

lim[x→5−] 1/((x−2)(x−5)) = 1/((3)(−0)) = −∞

lim[x→5+] 1/((x−2)(x−5)) = 1/((3)(+0)) = +∞

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- Mr.PersonaLv 53 years ago
I'll just give you the "by definition" things. In a general sense (and, consequently, in a topological sense) a limit would mean that as close as we get to f(x) (e.g. r distance from f(x)), we can find a y near x such that f(y) is within r distance of f(x). I.e., as close as we get to f(x), the function exists in that neighborhood.

Now, looking specifically at R, we can be more specific. Since a function only connects at two sides, i.e the left & right side, as long as from the left & right side we are approaching f(x), then the limit exists.

In short. You always have a left hand and a right hand limit. Why? Because for f(x), you can always considering sequences of points z_1, z_2, z_3,... < x as z_n get closer to x, and w_1 w_2, w_3, ... > x as w_n gets closer to x. I.e., these sequences are always well-defined. Given these sequences, you can always see what they tend towards (could be a value, could be positive infinity, could be negative infinity. You should know this from how sequences of parial sums can converge to a finite value or to infinity).

Then, only if these values agree (meaning the concept of a "limit" is well-defined in this instance due to what I state in my first paragraph. I.e., any location near f(x), which for R is just the left & right), does the limit exist & equal this value.

Note for limits where x->infinity there is no such thing as a right hand limit. Why? Because the limit exists with just the left hand limit. Why? Because there is not such concept as coming from the "right" of infinity.

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- Randy PLv 73 years ago
You know by analyzing which way the trend is going. The important thing to remember is that the limit is defined as what happens when you take a sequence of x values which APPROACH BUT DO NOT EQUAL the limiting value. It is very important to realize that you don't just plug in the final value, you look at values near it.

For instance, what is the limit of (x + 2)/(x - 3) as x approaches 3 from the left? Clearly when x = 3 the denominator is 0 so this ratio is going to blow up. But which way?

"Limit as 3 approaches from the left" means we're going to take a sequence of x values like 2.9, 2.99, 2.999, ... All of them are less than 3. All of them. So x - 3 is always negative. And x + 2 is always positive. Therefore this ratio is always negative for that sequence, and it is going toward -infinity.

Feel free to ask other examples where you can't understand how the limit was analyzed.

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