Subject: Calculus

Limit Of A Function

The direct application of the definition of a limit doesn't really allow us to easily solve any limit related problems. Because of this, we need to come up with some shortcuts to help us solve the limit of a function in a more straightforward manner. We've previously discussed that the limit of a function corresponds to the value of the function as it approaches a certain point. Here, we will try to find that value with the help of some simplified mathematical techniques. Theorems will be presented, and examples will be provided to help you better understand the concept. These are some important theorems your teacher will probably ask you to memorize. Take it slow, read through a few times and then try some of the examples!

The Ten Shortcuts

THEOREM 1 (A Limit is Unique)

\hbox{If }\lim_{x \to a} f(x) = L_{1} \hbox{ and } \lim_{x \to a} f{x} = L_{2 }\hbox{, then } L_{1}=L_{2}

This theorem only signifies the uniqueness of a limit of a function f(x) if it exists. In layman's terms, a specific function has a single specific limit as it approaches a specific point. Sounds confusing? Well, we're basically back in Algebra here (if a = b and a = c then b = c). You have to start somewhere right?

THEOREM 2 (Limit of a Linear Function)

If c and d are constants, then \lim_{x \to a}(cx+d)=ca+d.

Example #1

Solve the following limit:

\lim_{x \to 2}(4x-2)

Solution: Applying the above theorem where c=4 and d=2, we have

\lim_{x \to 2}(4x-2)=4(2)-2=8-2=6
.

THEOREM 3 (Limit of a Constant)

If c is a constant, then \lim_{x \to a}c=c for any real numbers a & c

Example #2

Find the limit of the following:

\lim_{x \to 9} \sqrt{6}

Solution: By the above theorem, it is clear that when c=\sqrt{6}, then:

\lim_{x \to 9} \sqrt{6} = \sqrt{6}

THEOREM 4 (Limit of an Identity Function)

For any real number c:

\lim_{x \to b}x=b

Example 3

Solve the following limit:

\lim_{x \to \sqrt{29}}x

Solution: By Theorem 4, it is clear that when b=\sqrt{29}, then:

\lim_{x \to \sqrt{29}}x = \sqrt{29}

THEOREM 5 (Limit of the Sum of a Function)

\hbox{If } \lim_{x \to a}f(x)=L \hbox{ and } \lim_{x \to a}g(x)=N

then,

\lim_{x \to a}[g(x)+f(x)]=L+N

Theorem 5 can be simplified to: the limit of a sum is equal to the sum of the respective limits if and only if the limit exists. Consider the following example.

Example #4

Solve the following limit:

\lim_{x \to \sqrt{29}}x + \lim_{x \to \sqrt{29}} \sqrt{6}

Solution: By Theorem 5, Example 2 and 3, it is clear that:

\lim_{x \to \sqrt{29}}x + \lim_{x \to 9} \sqrt{6} = \lim_{x \to \sqrt{29}} [x+\sqrt{6}] = \sqrt{29}+\sqrt{6}

THEOREM 6 (Limit of the n-number of a Function)

If,

\lim_{x \to a}f_{1}(x)=L_{1}, \lim_{x \to a}f_{2}(x)=L_{2},..., and \lim_{x \to a}f_{n}(x)=L_{n},

then:

\lim_{x \to a}[f_{1}(x)+f_{2}(x)+ \cdot \cdot \cdot +f_{n}(x)=\lim_{x \to a}f_{1}(x)+\lim_{x \to a}f_{2}(x)+ \cdot \cdot \cdot + \lim_{x \to a}f_{n}(x)

= L_{1} + L_{2}+ \cdot \cdot \cdot + L_{n}

Note: Theorem 6 is just corollary to Theorem 5. The former extends the number of functions to n



THEOREM 7 (Limit of a Product of a Function)

If,

\lim_{x \to a}f(x)=L_{1} \hbox{and } \lim_{x \to a}g(x)=L_{2}

then:

\lim_{x \to a}[f(x)+g(x)] = \lim_{x \to a}f(x) + \lim_{x \to a}g(x) = L_{1}+L_{2}

THEOREM 8 (The product of a constant and a limit)

If,

\lim_{x \to a}f(x) = L

and k is a constant, then:

\lim_{x \to a}kf(x) = k\lim_{x \to a}f(x) = kL

Similar to the limit of the sum of functions, the theorem on the limit of the product of functions may as well be extended to n number of functions provided that the limit of every function exist.

Example #5

Solve the following limit:

\lim_{x \to -2}[(7x+11)(-4x)]

Solution: By the above theorem, it can be easily be deduced that:

\lim_{x \to -2}[(7x+11)(-4x)] = \lim_{x \to -2}(7x+11)\lim_{x \to -2}(-4x)

Thus,

\lim_{x \to -2}[(7x+11)(-4x)] = -3 x 8=-24

THEOREM 9 (Limit of a Function with Power n)

If,

\lim_{x \to a}f(x)=L

and n is a positive integer, then we have:

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

Example #6

Solve the following limit:

\lim_{x \to 1}(x+11)^{6}

Solution: We can solved this by performing power multiplication six times but it will take much of our time. But by the above theorem, it can be seen directly that:

\lim_{x \to 1}(x+11)^{6} = [\lim_{x \to 1}(x+11)]^{6} = 12^{6}

THEOREM 10 (Limit of the Quotient of a Function)

If,

\lim_{x \to a}f(x) = L \hbox{ and } \lim_{x \to a}g(x)=N

then:

\lim_{x \to a}\frac{f(x)}{g(x)} = \frac{\lim_{x \to a}f(x)}{\lim_{x \to a}g(x)} = \frac{L}{N}

Example #7:

Solve the following limit:

\lim_{x \to 0}\frac{4x^{24}+1}{x+3}

Solution: With the help of Theorem 10, it can be easily solved as:

\lim_{x \to 0}\frac{4x^{24}+1}{x+3} = \frac{\lim_{x \to 0}4x^{24}+1}{\lim_{x \to 0}x+3} = \frac{1}{3}

Next Topic: One-sided Limit