Limits is one of the foundational topics in calculus that plays a vital role in solving real world problems in science and engineering. In the previous article, “Limits: An introduction and the basic methods to solving different limits”, we explored the primary methods used in solving the different types of limits you can encounter. To recap:

You can solve several limit exercises with the aid of the following three techniques:

Direct substitution

Assuming you have a limit like:

\(lim_{x \to \frac{\pi}{3}} \tan x\)

Simply substitute the given number into the variable to get the value √3

Factorization (usually for fractions)

If you realize directly subisuiting will give you an imaginary number then you might notice in your question that a certain line of expressions can be rewritten into a form whereas it will match with another in the denominator. Once you get that, the two expressions may cancel out and you are free to continue simplifying and finally substituting.

\(lim_{z \to 4} \frac{4 – z}{z^2 – 16}\)

\( z^2 – 16 \) is the same thing as \( z^2 – 4^2 \), so we can rewrite our denominator as \( (z – 4)(z + 4) \) based on the difference rule

= \( \frac{4 – z}{(z – 4)(z + 4)} \)

= \( \frac{4 – z}{-1(4 – z)(z + 4)} \)

\(lim_{z \to 4} \frac{1}{-(z + 4)} \)

= \( \frac{1}{-(4 + 4)} = -\frac{1}{8} \)

Conjugate (usually for radicals)

This method is used whenever you have a radical expression in your question. Multiplying a radical expression with it’s conjugate (switching the operator sign) will always cancel out the root, leaving you with a simpler expression to simplify.

\(lim_{x \to 6} \frac{\frac{1}{sqrt{x+3}} – \frac{1}{3}}{x – 6}\)

Start by multiplying the top and bottom by \( 3\sqrt{x+3} \) (the common denominator of the two fractions)

= \(lim_{x \to 6} \frac{3 – \sqrt{x+3}}{3(x-6)\sqrt{x+3}}\)

Multiply the top and bottom by \( 3 + \sqrt{x+3} \) (the conjugate of the radical)

= \(lim_{x to 6} \frac{9 + 3\sqrt{x+3} – 3\sqrt{x+3} – (x+3)}{3\sqrt{x+3}(x-6)(3 + \sqrt{x+3})}\)

Simplify:

= \( \frac{9 – x – 3}{3\sqrt{x+3}(x-6)(3 + \sqrt{x+3})} \)

= \(lim_{x \to 6} \frac{-1(x-6)}{3\sqrt{x+3}(x-6)(3 + \sqrt{x+3})}\)

= \(lim_{x \to 6} \frac{-1}{3\sqrt{x+3}(3 + \sqrt{6+3})}\)

Replace \( x \) with \( 6 \):

= \(-\frac{1}{54}\)

In this article, you will learn about limits at infinity, when a limit does not exist, and another powerful method for solving limits – a technique that often acts as a shortcut.

When does a limit not exist

Before we proceed, there is an important concept that should be clarified: What does it mean for a limit to not exist?

It is not simply when you get an imaginary number. A limit is said to not exist when a function behaves inconsistently as it approaches a certain point. In other words, the function does not approach a singular, definite value. There are several ways to investigate this condition:

Left-hand and right-hand limits are not equal

If the limit from the left (\(\left(x \to a^{-}\right)\)) is different from the limit from the right (\(\left(x \to a^{+}\right)\)), then the overall limit at that point does not exist.

Example: \(lim_{x \to 0} \frac{|x|}{x}\)

Function grows without bound

If the function increases or decreases indefinitely as it approaches a point, the limit is considered infinite and does not exist in the conventional sense.

Example: \(lim_{x\to 0} \frac{1}{x^2} \)

Oscillating behaviour

If the function continually oscillates between values as it approaches a point, the limit cannot settle on a single number.

Example: \(\lim_{x\to 0}\sin\frac{1}{x} \)

Jump discontinuities

If the function has a sudden jump at a point, the limit at that point does not exist because it does not approach a single value.

Example: A step function like the Heaviside function at the jump point.

Undefined function values at a point

If the function is undefined at a point and there’s no way to approach a single finite value through algebraic or limit techniques, the limit does not exist.

Limits at infinity

“The limit of the function \(\sqrt[x]{x}\) as x approaches infinity”

What does limit to infinity mean?

When you see a limit like that, it is describing what happens to a function as the x variable gets larger and larger – infinitely large (either in the positive or negative direction). Conceptually this is helpful because it allows us to know the long term behaviour of a function like if it grows without bound, or if it oscillates, or if it approaches a specific value, or behaves in another unpredictable way. This type of analysis is widely used in the world of science, engineering, and mathematical modeling.

But how do you EVALUATE such limits? The best way to understand this is through examples, as there are specific principles that govern limits at infinity.

Case 1

\(lim_{x \to \infty} x\)

Visualize this: as x gets huge, the graph of the above function hugs the x-axis more and more. It never touches it, but it approaches it.

Mathematically this will simply be equal to \(\infty\).

If you had: \(lim_{x \to \infty} x^3\)

Then the answer would still be infinity. Substituting infinity into x²will give you infinity², so it will just be a larger infinity which is still infinity.

What about?: \(lim_{x \to \infty} x^3\)

Answer: \(\infty\)

Case 2

\(lim_{x \to \infty} 3x\)

Substituting your \( x \) you will get \( = 3(\infty) \)

\( = \infty \)

Think about it this way: 3 times any number still gives you a number.

\( lim_{x \to \infty} (5 + 2x – x^3) \)

You could solve this step by step by substituting:

\(5 + 2(\infty) – (-\infty)^3 \)

\( = 5 + \infty – (-\infty) \)

\( = 5 + \infty + \infty \)

\( = \infty \)

Note that \( \infty – \infty \) does not equal zero. The same way \( \infty + \infty \) would not give you \( 2(\infty) \).

But an easier way to solve these types of polynomials is to cancel out the more insignificant terms like so:

Limit as \( x \to -\infty \) for \( 5 + 2x – x^3 \)

= \(lim_{x \to -\infty} (-x^3)\)

\( x^3 \) is the term with the highest power so the remaining can be disregarded.

So you’ll be left with \( -(-\infty)^3 \)

= -(-\infty)

= \(\infty\)

Case 3

\(lim_{x \to \infty} \frac{1}{x}\)

Let us take \( x \) as 1 to 9 and substitute it into the given function \( 1/x \):

\[\begin{aligned}\frac{1}{1} &= 1 \\\frac{1}{2} &= 0.5 \\\frac{1}{3} &= 0.33 \\\frac{1}{4} &= 0.25 \\\frac{1}{5} &= 0.2 \\\frac{1}{6} &= 0.16 \\\frac{1}{7} &= 0.14 \\\frac{1}{8} &= 0.125 \\\frac{1}{9} &= 0.11\end{aligned}\]

Do you notice the pattern? As our \( x \) gets larger and larger the function gets closer and closer to 0.

So in mathematics we say \( \frac{1}{\infty} = 0 \).

The same would apply for the left hand side: \( \frac{1}{-\infty} = 0 \).

Example 1

\(lim_{x \to -\infty} (3x^3 – 5x^4)\)

Discard the insignificant term and rewrite the limit as:

\( lim_{x \to -\infty} -5x^4 = -5(-\infty)^4 = -5(\infty) = -\infty \)

Example 2

\( lim_{x \to \infty} \frac{5x + 2}{7x – x^2} \)

There are two ways you can go about doing limits at infinity with rational functions:

The first method follows the same principle you learnt earlier with the polynomials.

Disregard the insignificant terms (both in the numerator and denominator) and only deal with the terms with the highest power.

Like so:

The numerator would go from \( 5x + 2 \) to \( 5x \).

The denominator would go from \( 7x – x^2 \) to \( -x^2 \).

So your limit is now:

The limit as \( x \) approaches \( -\infty \) for \( \frac{5x}{-x^2} \).

At this point the \( x \) at the top and bottom can cancel:

= \(\frac{-5}{x} \) = \(\frac{-5}{-\infty} \) = 0

Now for the second method; You will take the highest power in the denominator and multiply its reciprocal with the top and bottom part of the fraction.

In this case it is \( x^2 \).

Multiplying the top and bottom by \( 1/x^2 \), you will end up with:

The limit as \( x \) approaches \( -\infty \) for:

\( \frac{\frac{5}{x} + \frac{2}{x^2}}{\frac{7}{x} – 1} \)

Now substituting the infinity you will get:

\( \frac{0 + 0}{0 – 1} = \frac{0}{-1} = 0 \)

The first method is more direct and it’s usually an intuitive way of solving, but for those who aren’t used to these questions yet, the second method is probably easier.

Either works though!

Example 3

\( lim_{x \to \infty} \frac{8x^2 – 5x}{4x^2 + 7} \)

If you solve this using the first method you can instantly see that your question reduces to the limit as \( x \) approaches \( \infty \) for \( \frac{8x^2}{4x^2} = \frac{8}{4} \).

But let us follow the second method step by step so that you can familiarize yourself with it more:

Multiply the top and bottom by \( 1/x^2 \):

\( lim_{x \to \infty} \frac{8 – \frac{5}{x}}{4 + \frac{7}{x^2}} = \frac{8 + 0}{4 + 0} = \frac{8}{4} = 2 \)

L’Hopital’s rule – the shortcut technique

As seen previously, solving limits using algebraic manipulation usually requires several steps. However, in some cases you can use L’Hopital’s rule which allows you to evaluate limits involving indeterminate forms by differentiating the numerator and denominator separately.

L’Hopital’s rule provides a much quicker and efficient process to solving exercises

Example 1

\(lim_{z \to 4} \frac{4 – z}{z^2 – 16}\)

As we went through in recap, to solve this question, we must rewrite the denominator to \( (z-4)(z+4) \) correct? In order to get the final answer which is \( -\frac{1}{8} \). 

Now if we were to use L’Hôpital’s rule instead…

\(lim_{z \to 4} \frac{4 – z}{z^2 – 16} = lim_{z \to 4} \frac{-1}{2z} = -\frac{1}{2 \times 4} = -\frac{1}{8}\)

See how quickly that was solved? The L’Hôpital’s rule is a really great technique you can use when your time is limited or when you want to check your answer.

Example 2

\(lim_{x \to \infty} \frac{x^2}{e^x}\)

If you were to substitute infinity directly you would get:

\( \infty / \infty = \text{indeterminate} \)

So differentiate the top and bottom to get:

The limit as \( x \) approaches infinity for \( \frac{2x}{e^x} \)

Then differentiate again (yes, that’s allowed):

\( \frac{2}{e^x} \)

So the limit as \( x \) approaches \(\infty\) for \( \frac{2}{e^x} = 0 \)

Example 3

\(lim_{x \to \infty} \frac{\ln x}{x} = lim_{x \to \infty} \frac{1/x}{1} = \frac{1}{\infty} = 0\)

Example 4

\(lim_{x \to 0} \frac{sin(7x)}{sin(4x)}\)

\( sin(0)/sin(0) = 0/0 \) so you’d have to use L’Hôpital’s rule in this case too:

\(lim_{x \to 0} \frac{7cos(7x)}{4cos(4x)} = \frac{7(1)}{4(1)} = \frac{7}{4}\)

Practise questions

Try the following questions to put what you learned to the test:

\(lim_{x \to \infty} \frac{x^2 + 5x}{2x^2 – x + 1}\)

Answer: 1/2

\(lim_{x \to \infty} \left(\sqrt{x^2 + 3x} – x\right)\)

Answer: 3/2

\(lim_{x \to \infty} \frac{\ln x}{\sqrt{x}}\)

Answer: 0

\(lim_{x \to -\infty} \frac{2x^3 + x^2 – 1}{x^3 – 4x}\)

Answer: 2

\(lim_{x \to 0} \frac{e^x – 1}{x}\)

Answer: 1

\(lim_{x \to 0} \frac{1 – \cos x}{x^2}\)

Answer: 1/2

\(lim_{x \to \infty} \frac{x^3}{e^x}\)

Answer: 0

\(lim_{x \to \infty} \frac{(\ln x)^2}{x}\)

Answer: 0

\(lim_{x \to 0} \frac{\tan(5x)}{x}\)

Answer: 5