# Visualizing 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!

# 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.

So:

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

# Using the Definition of a Derivative…

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

f ‘ (x) = lim (as h goes to 0) of [f(x+h) – f(x)] / h

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

f(x+h) = 6sqrt(x+h)

f(x) = 6sqrt(x)

So, the derivative is:

Step 1:  The set-up)  f ‘ (x) = limit as h goes to 0 of [6sqrt(x+h) – 6sqrt(x)]/h

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

The general strategy to use when you see square roots is to multiply both top and bottom of the fraction by the conjugate, which is (6sqrt(x+h) + 6sqrt(x):

Step 2:  Simplifying)  [6sqrt(x+h) – 6sqrt(x)] * (6sqrt(x+h) + 6sqrt(x)  ) / h * (6sqrt(x+h) + 6sqrt(x)  )

Simplify this more and nice things happen (distribute / FOIL the top):

36(x+h) – 36(x) / [h *(6sqrt(x+h) + 6sqrt(x))]

Simplify numerator more:

(36x + 36h – 36x) / [h *(6sqrt(x+h) + 6sqrt(x))]

36h / [h *(6sqrt(x+h) + 6sqrt(x))]

Cancel out “h”

36 / (6sqrt(x+h) + 6sqrt(x))]

Now, take the limit as h goes to 0 (plug in 0 for h):

36 / (6sqrt(x+0) + 6sqrt(x))] =

36 / (6sqrt(x) + 6sqrt(x))

36 / 12 sqrt(x)