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Limits vs Values

Look at the function f(x) in orange below:


We are going to answer 4 questions about this graph.  They are all related to each other, but different questions.  Seeing the difference will help us sort out the difference between a function value and a limit.

Q1:  Find f(1)

Q2:  Find  limx→1 f(x)

Q3:  Find  limx→1+ f(x)

Q4:  Find  limx→1 f(x)

OK….. Try to answer these questions with what you know… Then continue reading to see the answers and explanations!

Continue reading Limits vs Values

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Basic Concept of a Limit

Here is a brief example on the concept of a limit:

Look at the function f(x) in orange below:


Question 1.  Find f(3)

Explanation of question 1: Find the value of the function when you plug in 3.  What is the height of the function at the exact moment when x=3?

Answer 1:  The function is undefined at x=3.  There is a hole when x=3.

So, f(3) is undefined.

Question 2:  Find limx→3f(x)

Explanation of question 2:  We are being asked to find what the function is doing around (but not at) 3.  What is happening to the path of the function on either side of 3?

In order to find limx→3f(x), we must confirm that limx→3+  f(x) and limx→3–  f(x) both exist and are equal to each other.

So, let’s find limx→3+  f(x).  What is happening to the function values as you approach x=3 from the right-hand side?  Literally run your finger along as if x=4, then x=3.5, then x=3.1.  What value is the function getting closer to?


The function is approaching a height of 4.

Let’s find limx→3–  f(x).  What is happening to the function values as you approach x=3 from the left-hand side?  Literally run your finger along as if x=1, then x=2, then x=2.9.  What value is the function getting closer to?


The function is also approaching a height of 4.


limx→3+  f(x) = 4

limx→3–  f(x) = 4

Since, the left-handed limit at 3 and right-handed limit at 3 exist and are equal, this gives:

limx→3f(x) = 4.

So, to summarize, here are 4 different things we found.  They are related, but not necessarily the same:

f(3) is undefined

limx→3+  f(x) = 4

limx→3–  f(x) = 4

limx→3f(x) = 4

Are you ready to try one on your own? Click here! (Don’t worry, I’ll walk you through the solutions too)

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Limit Example

Q:  What is limx→2 (x-2)/|x-2|

A:  This question is not too bad if you know what the function (x-2)/|x-2| looks like graphically.  But, let’s say you don’t.

We are going to “talk our way” through this problem to help solidify the concept of a limit.

If you plug in 2 to the function, you are finding the value of the function when x=2.  This is important, and related, though it is not the limit.  This is even sometimes a skill used to help us find the limit, but it is still not the limit.  Sometimes the function value is equal to the function limit, which can also be confusing, but not all the time.  Let’s find the value of the function when x=2:

(2-2)/|2-2| = 0/0 = undefined.

Okay.  This function is undefined when x=2.  This means there is a hole, or an asymptote, or a break or a jump or some disruption in the continuity of the function. Continue reading Limit Example

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One-sided Limit Example

Q:  Find the one-sided limit (if it exists):

limx→-1–     (x+1)/(x4-1)

A:  So we need to find the limit of this function (x+1)/(x4-1) as x approaches -1 from the left.  Remember, from the left means as x gets closer and closer to -1, but is still smaller.

The concept: What is happening to this function as x = -2, x = -1.5, x = -1.1, x = -1.0001, etc…

We test first and plug -1 into the function: (-1+1)/((-1)4-1) = 0/0

Whenever you get 0/0, that is your clue that maybe you need to do “more work” before just plugging in or jumping to conclusions.

So, let’s try “more work” — usually that means simplifying.  I see that the denominator can factor.  We have:

(x+1) / (x4-1) = (x+1) / [(x2-1)(x2+1)]

Let’s keep factoring the denominator:

(x+1) / [(x-1)(x+1)(x2+1)]

Now, it appears there is a “removable hole” in the function.  This means, we can remove this hole by reducing the matching term in the numerator with the matching term in the denominator:

(x+1) / [(x-1)(x+1)(x2+1)]

= 1 / [(x-1)(x2+1)]

Notice that hole exists when x = -1 (and it was removable! This is good news for us since we are concerned with the nature of the function as x approaches -1)

Continue reading One-sided Limit Example

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Finding the limit example

Q: Find the limit (as x approaches 3) of (x3 – 27) / (3-x)

A:  The first thing to do when finding a basic limit is try plugging in the number in question (3).

So, plug in 3 to get:

(33 – 27) / (3-3) = 0/0 <– if you get 0/0 or infinity/infinity that means there is more work to be done.  However, if you had just got a number like 4 or something, that would’ve been your answer!

OK, we got 0/0 so that means more work.  More work could mean many things (apply different rules, factor and cancel, simplify, etc).  In this case, it appears we can factor, so we try that:

(x3 – 27) / (3-x) = (x – 3)(x2 + 3x + 9)/ (3 – x)

Now, here comes some tricky insight.  I notice that the (x – 3) on top is very similar to the (3 – x) on the bottom.  I am going to factor a “-1” out of the (x – 3) that is on top.

Notice:  -1(3 – x) = (x – 3)

So, the numerator becomes:

-1(x – 3)(x2 + 3x + 9)/ (3 – x)

Now, the (x – 3) term cancels from the top and bottom to leave:

-1(x2 + 3x + 9)

So, we are trying to find the limit (as x approaches 3) of -1(x2 + 3x + 9).  We have “removed the hole” — the factor (3 – x) was a “removable hole” that was causing calculation problems.  In the simplified version, we can plug in the value 3 to now calculate where that hole was occuring:

Lim (x –> 3) of -1(x2 + 3x + 9) = -1(32 + 3(3) + 9) = -1(9 + 9 + 9) = -27

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Using the Definition of a Derivative

Q:  Use the definition of a derivative to take the derivative of f(x) = 6sqrt(x)

Answer:  First, let’s start with the definition of a derivative:

f ‘ (x) = lim h→0 [f(x+h) – f(x)] / h

So, using our given equation:  f(x) = 6sqrt(x), we have:

f(x+h) = 6√(x+h)

f(x) = 6√(x)

So, the derivative is:

Step 1:  The set-up)  f ‘ (x) = limit h→0 of [6√(x+h) – 6√(x)]/h

Now, we need to do some simplifying before plugging in 0 for h.

Continue reading Using the Definition of a Derivative