In calculus, a limit describes what becomes of a function as it reaches a certain value. For example:
\(\lim_{x \to 5} 3x^2\)
It is read as “limit of \( 3x^2 \) as \( x \) approaches 5”
This basically means what value does the function \( 3x^2 \) get closer to as \( x \) gets closer to 5? Mathematically, if we substitute \( x \) with 5 directly, we get 75. But the actual idea of a limit is that the function will not actually reach that exact value. It will only approach it infinitely closely.
Limits are essential in engineering, medicine, and physics. They help describe how systems behave as conditions approach a certain point. For example, how velocity changes at an instant or how a reaction rate behaves near equilibrium.
In this article students will learn how to solve limits using:
- Direct method (Substitution)
- Conjugate method
- Factorization method
But before that, here are some general limit rules you’ll need to know. It is straightforward laws that explain the result of different operations applied to two distinct limits.
*Table by Mathtuition88.com
Substitution
Example one
\(\lim_{x \to 2} \frac{x^2 – 4}{x + 2} \)
When faced with a limit question, always the first thing to do is to substitute the given number in place of the variable.
So for the limit of \(\frac{x^2-4}{x+2}\) as \( x \) approaches -1 you get \( (-1)^2-4)/((-1)+2) = 1-4/1 = -3\)
Example two
\(\lim_{x \to 5} \frac{x^2 – 3x + 4}{x + 5} \)
Substitute 5 into all the places x was and you get \((5)^2-3(5)+4)/((5)+5) =(25-15+4)/(10) = 14/10 = 7/5\)
Example three
\(\lim_{x \to 0} \frac{5x}{x^2 + 2x} \)
For this one let’s first rewrite the denominator as \( x(x+2) \)
So our function now is \(\frac{5x}{x(x+2)}\)
The x at the top and the x at the bottom will cancel each other out leaving you with \(\frac{5}{(x+2)}\)
\(=5/((0)+2) = 5/2\)
Factorization
Example one
\(\lim_{x \to 2} \frac{x^2 + 3x – 10}{x – 2} \)
As usual, substitute your number into the variable. That will be: \(((2)^2+3(2)-10)/((2)-2) = (4+6-10)/(0) = 0/0\)
As you can see you will get 0/0 which in mathematics is not a real number. You cannot divide by zero so you can see that here we have reached some sort of roadblock.
You will need to go back to the function and see how you can manipulate or simplify to get an expression where our final result will be a real number.
\( x^2+3x-10 \) is actually a quadratic trinomial so it can be easily broken down to \((x+5)\) and \((x-2)\).
The function becomes \(\frac{((x+5)(x-2))}{(x-2)}\)
Notice that now we have two identical expressions that will cancel each other out leaving us with:
\((X+5)/1 = (2)+5 = 7\)
Example two
\(\lim_{x \to 3} \frac{x^2 – 9}{x – 3} \)
If we replace x with 3 we get:
\(((3)^2-9)/((3)-3) \)
\(=(9-9)/0 =0/0\)
Once again we have another limit question that requires an extra step. If you look closely \( x^2-9 \) can be opened up. It is difference of two square and recall that:
Difference of squares – \( a^2 – b^2 = (a – b)(a + b) \)
Difference of cubes – \( a^3 – b^3 = (a – b)(a^2 + ab + b^2) \)
where a is x and b is \( 3^2 \)
So we can rewrite the function like so:
\(\frac{((X+3)(X-3))}{(X-3)}\)
X-3 at the denominator and numerator cancel out each other so \( x+3 = (3)+3 \) and our final answer is 6
Example three
\(\lim_{x \to 2} \frac{x^3 – 8}{x – 2} \)
At this point you should be noticing the patterns, so right off the bat without even solving the complete question you can see that if we substitute 2 we get 2-2=0 in the denominator. Looking at the numerator we can see that we have a difference of cubes where a is \( x^3 \) and b is \( 2^3 \) hence our function can be rewritten as:
\(\frac{((X-2)(X^2+2x+4))}{(X-2)} \)
\(=X^2+2x+4/1\)
\(=(2)^2+2x+4\)
\(=4+4+4 = 4\times3=12\)
Conjugate
Example one
\(\lim_{x \to 4} \frac{\frac{1}{x} – \frac{1}{2}}{x – 4} \)
Whenever you have a radical in a limit what you need to do to get rid of it is to multiply by it’s conjugate.
So apply that to the question given the conjugate would be \(\sqrt{x} +2 \)
Whatever you do in the numerator you must reciprocate in the denominator and visa versa so:
Lim \( x\to4 \) \(( \sqrt{x} -2)\times(\sqrt{x} +2)/(x-4)\times(\sqrt{x} +2)\)
\(= (X+2\sqrt{x} -2\sqrt{x} -4)/(X-4)(\sqrt{x} +2)\)
\(=(X-4)/(X-4)(\sqrt{x} +2)\)
\(=1/(\sqrt{x}+2)\)
\(=1/(\sqrt{4} +2) =1/(2+2) \)
\(= 1/4\)
Example two
\(\lim_{x \to 3} \frac{\frac{1}{x} – \frac{1}{3}}{x – 3} \)
The limit here looks a bit more complicated but you are to follow the same steps from the first example
Lim \( x\to0 \) \(( \sqrt{(x+2)}-\sqrt{2})\times(\sqrt{(x+2)}+\sqrt{2})/(x)\times(\sqrt{(x+2)}+\sqrt{2})\)
\(=(X+2+\sqrt{(x+2)}-\sqrt{2} – \sqrt{(x+2)}-\sqrt{2} -2)/(X(\sqrt{(x+2)}+\sqrt{2}))\)
\(=X/(x(\sqrt{(x+2)}+\sqrt{2}))\)
\(=1/(\sqrt{(x+2)}+\sqrt{2})\)
\(=1/(\sqrt{(0+2)} +\sqrt{2})\)
\(=1/(2\sqrt{2})\)
Example three
\(\lim_{x \to 0} \frac{\sqrt{x + 2} – \sqrt{2}}{x} \)
If you get a limit like this the simplest thing you can do to get rid of the fractions is multiply the numerator and denominator by the (fraction at the top) common denominator
Between \(\frac{1}{x}\) and \(\frac{1}{3}\) the common denominator is \(3x\) so:
Limit -> \(((\frac{1}{x} – \frac{1}{3})\times(3x))/((x-3)\times(3x))\)
\(=(3-x)/((3x)(X-3)\)
Now we are almost done but we have two expressions that are very similar yet not identical at the top and bottom.
If you factor out -1 for the expression 3-x (-1 is it’s GCF – greatest common factor) you can get x-3
Hence the function becomes:
\((-1(x-3))/((X-3)(3x))\)
\(=-1/3x = -\frac{1}{3}(3)\)
\(= -1/9\)
Example four
\(\lim_{x \to 4} \frac{\frac{1}{x} – \frac{1}{2}}{x – 4} \)
This is a rational fraction with both a fraction and a radical. Don’t panic. All we have to do is perform the methods for each individual type of limits question – multiplying the function by the conjugate AND multiplying the function by the common denominator. The order does not matter but let’s start by getting rid of the fraction:
\(\lim_{x \to 4} \frac{\left(\frac{1}{\sqrt{x}} – \frac{1}{2}\right) \times (2\sqrt{x})}{(x-4) \times (2\sqrt{x})} \)
= \(\frac{2-\sqrt{x}}{(x-4)(2\sqrt{x})}\)
Now multiply the top and bottom by the conjugate of the radical expression
= \(\frac{(2-\sqrt{x}) \times (2+\sqrt{x})}{((x-4)(2\sqrt{x}) \times (2+\sqrt{x})}\)
= \(\frac{4-x}{2\sqrt{x}(x-4)(2+\sqrt{x})}\)
Now simplify
= \(\frac{-1(x-4)}{2\sqrt{x}(x-4)(2+\sqrt{x})}\)
= \(\frac{-1}{2\sqrt{x}(2+\sqrt{x})}\)
= \(\frac{-1}{2\sqrt{4}(2+\sqrt{4})} = \frac{-1}{2(2)(2+2)} = \frac{-1}{4(4)}\)
And just like that the final answer is \( -1/16 \)!
Limits is one of the foundations in calculus that plays a vital role in solving real world problems in science and engineering. Even when direct substitution doesn’t work, by using methods such as substitution, factorization, and the conjugate approach, you can simplify complex expressions and determine the value of almost any example. In the next part of this essay, we’ll dive into more advanced examples and explore topics like L’Hopital’s rule and limits at infinity.
Practice questions
The following are more limit examples you can try yourself to practice using all the methods taught above:
- \(\lim_{x \to 1} \frac{8x + 2}{2x – 5} \)
Answer: -10/3
- \(\lim_{x \to \frac{\pi}{3}} \tan x \)
Answer: √3
- \(\lim_{x \to 3} \frac{\sqrt{12 – x} – 3}{\sqrt{7 – x} – 2} \)
Answer: 2/3
- \(\lim_{x \to 6} \frac{\frac{1}{\sqrt{x + 3}} – \frac{1}{3}}{x – 6} \)
Answer: -1/54
- \(\lim_{x \to 1} \frac{2x^2 – x – 3}{x + 1} \)
Answer: -1
- \(\lim_{x \to 4} \frac{4 – x^2}{x – 16} \)
Answer: 1
- \(\lim_{x \to 2} \frac{\sqrt{x + 7}}{x + 1} \)
Answer: 1
