(L3) Limits using Algebraic Methods#

In this lesson, we will see how to:

  • calculate the limit of a function at a point using algebriac methods

Lecture Videos#

Two Basic Limits#

Constant Function#

A constant function is any function where the output is the same constant \(C\) regardless of input \(x\). Specifically \(f(x)=C\) for all \(x\)-values.

The limit of a constant is then easy to calculate:

\[ \lim_{x\to a} C=C \]

As input \(x\) changes (\(x\to a\)) the output \(f(x)\) remains the same (it stays at that constant value \(C\)). If the input is changing, the constant function stays constant, it doesn’t change.

Identity Function#

The identity function is the function that just returns whatever the input value is. Specifically \(f(x)=x\) for all \(x\)-values.

The limit of the identity is also easy to calculate:

\[ \lim_{x\to a} x=a \]

As input \(x\) changes (\(x\to a\)) the output \(f(x)=x\) undergoes the exact same change: \(f(x)\to a\).

Limit Laws#

Limit Laws

(1) Constant Multiple \(\qquad \displaystyle \lim_{x\to a} k\cdot f(x) = k\cdot \lim_{x\to a} f(x) \)

(2) Limit of a Power \(\qquad \displaystyle \lim_{x\to a} \big[f(x)\big]^r = \big[\lim_{x\to a} f(x)\big]^r \)

(3) Limit of a Sum \(\qquad \quad \displaystyle \lim_{x\to a} f(x) + g(x) = \lim_{x\to a} f(x) + \lim_{x\to a} g(x)\)

(4) Limit of a Difference \(\quad \displaystyle \lim_{x\to a} f(x) - g(x) = \lim_{x\to a} f(x) - \lim_{x\to a} g(x)\)

(5) Limit of a Product \(\qquad \displaystyle \lim_{x\to a} f(x) \cdot g(x) = \lim_{x\to a} f(x) \cdot \lim_{x\to a} g(x)\)

(6) Limit of a Quotient \(\qquad \displaystyle \lim_{x\to a} \dfrac{f(x)}{g(x)} = \dfrac{\displaystyle \lim_{x\to a} f(x) }{\displaystyle \lim_{x\to a} g(x)}\)

Comments#

All of these Limit Laws only work provided:

  • the limits \(\displaystyle \lim_{x\to a} f(x)\) and \(\displaystyle \lim_{x\to a} g(x)\) both exist where applicable.

The Power Limit Law works provided:

  • exponent \(r\) is a positive integer or

  • exponent \(r\) is a root function of the form \(1/n\) where \(n\) is a positive integer. If \(r\) is an even root function (like square root \(r=1/2\)), then we also require \(f(x)>0\) near \(x=a\).

The Quotient Limit Law works provided:

  • the limit of the term in the denominator does not equal 0. Specifically we need \(\displaystyle \lim_{x\to a} g(x)\neq 0\) in order to apply this quotient law.

Example 1#

Compute the following limits algebraically:

(a) \(\displaystyle \lim_{x\to 2} x^3\)

We have a power function here, so we are going to use the Power Limit Law to get:

\[ \lim_{x\to 2} x^3 = \bigg(\lim_{x\to 2} x\bigg)^3 = \bigg(2\bigg)^3 \]

Notice that in that last step we are using the basic limit of the identity function:

\[ \lim_{x\to 2} x =2 \]
(b) \(\displaystyle \lim_{x\to 2} 5x^3\)

We have a constant multiple here, so we are going to use the Constant-Multiple Limit Law to get:

\[ \lim_{x\to 2} 5x^3 = 5\cdot \lim_{x\to 2} x^3 = 5\cdot 8 \]

Notice that in that last step we are using our previous result from part (a).

(c) \(\displaystyle \lim_{x\to 2} \bigg( 5x^3-15 \bigg)\)

We have some subtraction here, so we are going to use the Difference Limit Law to get:

\[ \lim_{x\to 2} \bigg( 5x^3-15 \bigg) = \lim_{x\to 2} 5x^3 - \lim_{x\to 2} 15 = 40-15 \]

Notice that in that last step we are using our previous result from part (b) and the result that the limit of a constant function is the constant (in this case \(15\)).

(d) \(\displaystyle \lim_{x\to 2} \bigg( 5x^3-15 \bigg)\cdot x^2\)

Now we have some multiplication of variable terms here, so we are going to use the Product Limit Law to get:

\[ \lim_{x\to 2} \bigg( 5x^3-15 \bigg)\cdot x^2= \lim_{x\to 2} \bigg( 5x^3-15 \bigg)\cdot \lim_{x\to 2}x^2 \]

We’ve already calculated the first limit, but we need to use the law again for power functions to get:

\[ = 25\cdot \big(\lim_{x\to 2}x\big)^2 = 25 \cdot \big(2\big)^2 \]
(e) \(\displaystyle \lim_{x\to 2} \dfrac{\bigg( 5x^3-2 \bigg)\cdot x^2}{x^3}\)

In this last one we have some division, so we need to use the Quotient Limit Law to get:

\[ \lim_{x\to 2} \dfrac{\bigg( 5x^3-2 \bigg)\cdot x^2}{x^3} = \dfrac{\displaystyle \lim_{x\to 2} \bigg( 5x^3-2 \bigg)\cdot x^2}{\displaystyle \lim_{x\to 2} x^3} = \dfrac{100}{8} \]

It is important to note that we are able to use this Quotient Limit Law because the limit of the terms in the denominator is not equal to \(0\).

Example 2#

Evaluate the following limit, justifying each step with the appropriate limit law.

\[ \lim_{x\to 3} \dfrac{x^2+4x-1}{2x-8} \]

Click through the tabs to see the steps of our solution.

We begin by noticing that this is a quotient, so we try applying the Quotient Limit Law (6).

\[ \lim_{x\to 3} \dfrac{x^2+4x-1}{2x-8} = \dfrac{ \displaystyle \lim_{x\to 3} \big(x^2+4x-1\big)}{ \displaystyle \lim_{x\to 3} \big(2x-8\big)} \]

Important - we need to keep this step in mind. As we go through the rest of our work, if we find that the limit of the terms in the denominator are \(0\), then this step is incorrect and we need to come back and revise our work.

We next apply the Sum and Difference Dimit Laws (3) and (4) to split up the limits:

\[\begin{split} &= \dfrac{ \displaystyle \lim_{x\to 3} \big(x^2+4x-1\big)}{ \displaystyle \lim_{x\to 3} \big(2x-8\big)} \\ & \\ &= \dfrac{ \displaystyle \lim_{x\to 3} x^2+ \lim_{x\to 3} 4x-\lim_{x\to 3} 1}{ \displaystyle \lim_{x\to 3} 2x-\lim_{x\to 3}8} \end{split}\]

The Power Law (2) is then used to bring the limit sign inside the exponent:

\[ = \dfrac{ \displaystyle \big(\lim_{x\to 3} x \big)^2+ \lim_{x\to 3} 4x-\lim_{x\to 3} 1}{ \displaystyle \lim_{x\to 3} 2x-\lim_{x\to 3}8} \]

Now we need the Constant Multiple Law (1) to bring the constant multiples to the front of the limit signs:

\[ = \dfrac{ \displaystyle \big(\lim_{x\to 3} x \big)^2+ 4\cdot \lim_{x\to 3} x-\lim_{x\to 3} 1}{ \displaystyle 2 \cdot \lim_{x\to 3} x-\lim_{x\to 3}8} \]

Finally, all of the limits that we remain to be calculated are of our 2 basic forms (constant and identity functions).

\[\begin{split} &= \dfrac{ \displaystyle \big(\lim_{x\to 3} x \big)^2+ 4\cdot \lim_{x\to 3} x-\lim_{x\to 3} 1}{ \displaystyle 2 \cdot \lim_{x\to 3} x-\lim_{x\to 3}8} \\ & \\ &= \dfrac{(3)^2+4\cdot(3)-1}{2\cdot(3)-8} = \dfrac{20}{-2} \end{split}\]

This is our answer! And notice that since the limit of denominator is not \(0\), we were correct to use the Quotient Limit Law in our first step.

Direct Evaluation#

Direct Evaluation

If \(f\) is a polynomial or rational function then:

\[ \lim_{x\to a} f(x) =f (a) \]

provided \(x=a\) is in the domain of \(f\).

Note:

  1. For polynomials \(p(x)\) this direct evaluation result holds for any \(x=a\).

  2. For rational functions \(\dfrac{p(x)}{q(x)}\) direct evaluation works as long as \(q(a)\neq 0\). This means we need to be careful with fractions and check that the denominator is non-zero when we plug-in.

Previous Example#

For the function \(f\) in the previous example, we calculated the limit (in exhausting detail) to be:

\[ \lim_{x\to3 } f(x)= \lim_{x\to3 } \dfrac{x^2+4x-1}{2x-8} = -10 \]

But since this is a rational function which is only undefined at \(x=4\), we see that direct evaluation actually yields the same result:

\[ f(3) = \dfrac{3^2+4(3)-1}{2(3)-8}=\dfrac{9+12-1}{6-8}=\dfrac{20}{-2}=-10 \]

So for this rational function we have:

\[ \lim_{x\to3 } f(x)= f(3) \]

First Strategy for Calculating a Limit#

This result actually gives us our first major strategy for calculating a limit simply:

  • Evaluate the polynomial or rational function directly at the \(x=a\) (the number \(x\) is approaching)

  • If the result is a number (no division by \(0\)), then that is our answer - we are done!

Example 3#

Compute the following limit:

\[ \lim_{x\to 0}\dfrac{x^2-4}{x-2} \]

Click through the tabs to see the steps of our solution.

We begin by recognizing that our function is:

  • a rational function and

  • \(x=0\) is in the domain (the number \(x\) is approaching)

Since this is a rational function which is defined at \(x=0\), we can calculate this limit by just evaluating the function:

\[ \lim_{x\to 0}\dfrac{x^2-4}{x-2} = \dfrac{(0)^2-4}{0-2}=\dfrac{-4}{-2}=2 \]

Example 8#

Compute \(\displaystyle \lim_{x\to 20} f(x)\) where \(f\) is a piecewise-function given by:

\[\begin{split} f(x)=\begin{cases} 100+12x & \quad \text{if} \quad x \leq 20 \\ \sqrt{180-4x} & \quad \text{if} \quad x > 20\\ \end{cases} \end{split}\]

Click through the tabs to see the steps of our solution.

In order to calculate this overall limit, we need to first calculate both of the one-sided limits. We usually need to do this for piecewise-functions if the limit we want to calculate happens to be near where the different pieces come together (in this case \(x=20\)).

Notice that left of \(x=20\) corresponds with the inequality \(x<20\) (smaller values). This will help us identify which piece to use for \(f(x)\):

\[ \lim_{x\to20^-} f(x)= \lim_{x\to20^-} (100+12x) = 100+12(20)=340 \]

Direct evaluation works in this last step since \((100+12x)\) is a polynomial.

Right of \(x=20\) corresponds with the inequality \(x>20\) (larger values), so we will need to use the square root piece of our function.

\[\begin{split} \lim_{x\to20^+} f(x)& = \lim_{x\to20^+} \sqrt{180-4x} \\ & = \sqrt{\lim_{x\to20^+}(180-4x)} \\ & = \sqrt{180-4(20)} \\ & = \sqrt{100} \\ & =10\\ \end{split}\]

Notice that we had to use the Power Limit Law first. This allowed us to bring the limit inside the square root, which then allowed us to use direct evaluation on the polynomial \(180-4x\).

We calculated the left-sided limit:

\[ \lim_{x\to20^-} f(x)= 340 \]

and the right-sided limit:

\[ \lim_{x\to20^+} f(x)= 10 \]

Since these limits are different, we can conclude that the overall limit does not exist. That is:

\[ \lim_{x\to20} f(x) \quad \text{DNE} \]

(Remember that both the left and right sided limits need to be equal to each other in order for the overall limit to exist.)