Notation

Notation

We can now introduce new notation. For the function \(y=\cfrac{(x+6)(x-2)}{x+6}\), we can write:

\[\lim_{x\to -\text{6}} \cfrac{(x+6)(x-2)}{x+6}=-\text{8}.\]

This is read: the limit of \(\cfrac{(x+6)(x-2)}{x+6}\) as \(x\) tends to \(-\text{6}\) (from both the left and the right) is equal to \(-\text{8}\).

Optional Investigation

Limits

If \(f(x)=x+1\), determine:

\(f(-\text{0.1})\)

\(\phantom{xxxxxx}\)

\(f(-\text{0.05})\)

 

\(f(-\text{0.04})\)

 

\(f(-\text{0.03})\)

 

\(f(-\text{0.02})\)

 

\(f(-\text{0.01})\)

 

\(f(\text{0.00})\)

 

\(f(\text{0.01})\)

 

\(f(\text{0.02})\)

 

\(f(\text{0.03})\)

 

\(f(\text{0.04})\)

 

\(f(\text{0.05})\)

 

\(f(\text{0.1})\)

 

What do you notice about the value of \(f(x)\) as \(x\) gets closer and closer to \(\text{0}\)?

Example

Question

Write the following using limit notation: as \(x\) gets close to \(\text{1}\), the value of the function \(y=x+2\) approaches \(\text{3}\).

This is written as:

\[\lim_{x\to 1}(x+2)=3\]

This is illustrated in the diagram below:

7219359c58f5587443e22d20e1829011.png

We can also have the situation where a function tends to a different limit depending on whether \(x\) approaches from the left or the right.

d613dc2e699631675cfe2208e608fb5e.png

As \(x\to 0\) from the left, \(f(x)\) approaches \(-\text{2}\). As \(x\to 0\) from the right, \(f(x)\) approaches \(\text{2}\).

The limit for \(x\) approaching \(\text{0}\) from the left is:

\[\lim_{x\to {0}^{-}}f(x)= -\text{2}\]

and for \(x\) approaching \(\text{0}\) from the right:

\[\lim_{x\to {0}^{+}}f(x)= -\text{2}\]

where \(0^{-}\) means \(x\) approaches zero from the left and \(0^{+}\) means \(x\) approaches zero from the right.

Therefore, since \(f(x)\) does not approach the same value from both sides, we can conclude that the limit as \(x\) tends to zero does not exist.

d1ba6264a8ec511acdfe7aac91a514da.png

As \(x\) tends to \(\text{0}\) from the left, the function approaches \(\text{2}\) and as \(x\) tends to \(\text{0}\) from the right, the function approaches \(\text{2}\). Since the function approaches the same value from both sides, the limit as \(x\) tends to \(\text{0}\) exists and is equal to \(\text{2}\).

Example

Question

Determine:

  1. \(\displaystyle\lim_{x\to 1}10\)
  2. \(\displaystyle\lim_{x\to 2}(x + 4)\)

Illustrate answers graphically.

Simplify the expression and cancel all common terms

We cannot simplify further and there are no terms to cancel.

Calculate the limit

  1. \(\displaystyle\lim_{x\to 1}10=10\)
  2. \(\displaystyle\lim_{x\to 2} (x + 4) = 2 + 4 = 6\)

f5e48dfe94308d597d223cd8e2a1e7b1.pngeed5423aa155242ef665a5333a9d9b11.png

Example

Question

Determine the following and illustrate the answer graphically:

\[\lim_{x\to 10}\cfrac{{x}^{2}-100}{x-10}\]

Simplify the expression

Factorise the numerator:

\(\cfrac{{x}^{2}-100}{x-10}=\cfrac{(x+10)(x-10)}{x-10}\)

As \(x \to 10\), the denominator \((x – 10) \to 0\), therefore the expression is not defined for \(x=10\) since division by zero is not permitted.

Cancel all common terms

\[\cfrac{(x+10)(x-10)}{x-10}=x+10\]

Calculate the limit

\begin{align*} \lim_{x\to 10}\cfrac{{x}^{2}-100}{x-10} &=\lim_{x\to 10} (x + 10) \\ &= 10 + 10 \\ &= 20 \end{align*}

Draw the graph

00121068f2542db53f57543973800c70.png

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This is a lesson from the tutorial, Differential Calculus and you are encouraged to log in or register, so that you can track your progress.

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