M2 Chapter: Limits - Your Comprehensive Study Notes
Hey everyone! Welcome to the foundational topic of Calculus: Limits. Don't worry if the word sounds a bit abstract. The goal of these notes is to make the concept of limits super clear and easy to understand. Limits are like the first, most important building block for everything else in Calculus, including differentiation and integration. So, let's get started and build a strong foundation together!
1. What on Earth is a Limit? (The Intuitive Idea)
Imagine you're walking towards a wall. You take steps that are half the remaining distance. You walk halfway... then half of the remaining way... then half of that remaining way... and so on. You get incredibly, ridiculously close to the wall, but you never technically touch it.
In mathematics, a limit is the value that a function "approaches" as the input (usually x) gets closer and closer to some number.
The key idea is that we don't care what the function's value is exactly at the number. We only care about what value it's heading towards from both the left and the right side.
The Notation
We write the limit of a function like this:
$$ \lim_{x \to a} f(x) = L $$Let's break that down:
- lim is short for "limit".
- x → a means "as x approaches the value a".
- f(x) is our function.
- L is the value the function is approaching.
So, the whole expression reads: "The limit of the function f(x) as x approaches a is L."
A Crucial Point: The Limit vs. The Actual Value
Sometimes, the limit at a point `a` is the same as `f(a)`. But not always! This is a super important concept.
Example: Consider the function $$ f(x) = \frac{x^2 - 4}{x - 2} $$.
What happens at `x = 2`? If you plug it in, you get $$ \frac{4-4}{2-2} = \frac{0}{0} $$, which is undefined! There's a "hole" in the graph at `x = 2`.
But what value does the function approach as `x` gets really close to 2?
Let's try values near 2:
f(1.9) = 3.9
f(1.99) = 3.99
f(2.01) = 4.01
f(2.1) = 4.1
See? The closer `x` gets to 2, the closer `f(x)` gets to 4. So, we say:
$$ \lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4 $$Even though `f(2)` is undefined, the limit as `x` approaches 2 exists and is equal to 4.
Key Takeaway: Section 1
A limit is the target value a function is heading towards as its input gets closer to a certain number. It doesn't matter if the function ever actually reaches that value.
2. How to Find Limits: The Toolkit
Okay, we can't just plug in numbers close to `a` every time. We need faster methods! Here are the main techniques you'll use.
Method 1: Direct Substitution (The Easy Way)
For most "nice" functions (like polynomials and many rational functions), the first thing you should always try is just plugging the value `a` directly into the function.
Example 1: Find $$ \lim_{x \to 3} (2x^2 - 5x + 1) $$
This is a polynomial, so it's "well-behaved". Just substitute `x = 3`:
$$ 2(3)^2 - 5(3) + 1 = 2(9) - 15 + 1 = 18 - 15 + 1 = 4 $$So, $$ \lim_{x \to 3} (2x^2 - 5x + 1) = 4 $$. Easy!
Method 2: When Direct Substitution Fails (The 0/0 Problem)
If direct substitution gives you an indeterminate form like $$ \frac{0}{0} $$, it doesn't mean the limit doesn't exist! It just means you need to do more work. This is a signal to simplify the function algebraically.
Technique A: Factorisation and Cancellation
This is what we could have used for our first example!
Example: Find $$ \lim_{x \to 2} \frac{x^2 - 4}{x - 2} $$
- Try Direct Substitution: We get $$ \frac{2^2 - 4}{2 - 2} = \frac{0}{0} $$. Time for algebra!
- Factor: The numerator is a difference of squares. $$ x^2 - 4 = (x-2)(x+2) $$.
- Rewrite and Cancel: $$ \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} $$ Since `x` is only approaching 2 (so `x ≠ 2`), the `(x-2)` term is not zero, and we can safely cancel it. $$ \lim_{x \to 2} (x+2) $$
- Use Direct Substitution Now: $$ 2 + 2 = 4 $$. The limit is 4.
Technique B: Rationalisation (Using the Conjugate)
Use this technique when you see a square root and get the $$ \frac{0}{0} $$ form.
Example: Find $$ \lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} $$
- Try Direct Substitution: $$ \frac{\sqrt{0+1} - 1}{0} = \frac{1-1}{0} = \frac{0}{0} $$. Algebra time!
- Multiply by the Conjugate: The conjugate of $$ \sqrt{x+1} - 1 $$ is $$ \sqrt{x+1} + 1 $$. Multiply the top and bottom by this. $$ \lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} \times \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1} $$
- Simplify: Remember that $$ (a-b)(a+b) = a^2 - b^2 $$. $$ \lim_{x \to 0} \frac{(\sqrt{x+1})^2 - 1^2}{x(\sqrt{x+1} + 1)} = \lim_{x \to 0} \frac{(x+1) - 1}{x(\sqrt{x+1} + 1)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x+1} + 1)} $$
- Cancel: Cancel the `x` from the top and bottom. $$ \lim_{x \to 0} \frac{1}{\sqrt{x+1} + 1} $$
- Use Direct Substitution Now: $$ \frac{1}{\sqrt{0+1} + 1} = \frac{1}{1+1} = \frac{1}{2} $$. The limit is 1/2.
Quick Review Box
Your Limit-Finding Strategy so far:
- Always try Direct Substitution first.
- If you get $$ \frac{0}{0} $$, it's a signal to do more work.
- Look for ways to simplify algebraically:
- Can you factor the top or bottom?
- Is there a square root? Try multiplying by the conjugate.
- After simplifying, try Direct Substitution again!
3. Theorems on Limits (The Rules of the Game)
Just like in algebra, we have rules that let us break down complicated limits into simpler ones. The syllabus requires you to recognise these, but thankfully, you do not need to prove them! They are very intuitive.
Let's assume that $$ \lim_{x \to a} f(x) = L $$ and $$ \lim_{x \to a} g(x) = M $$.
- Sum/Difference Rule: The limit of a sum is the sum of the limits.
$$ \lim_{x \to a} [f(x) \pm g(x)] = L \pm M $$ - Scalar Multiplication Rule: You can pull a constant out in front.
$$ \lim_{x \to a} [k \cdot f(x)] = k \cdot L $$ (where k is a constant) - Product Rule: The limit of a product is the product of the limits.
$$ \lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M $$ - Quotient Rule: The limit of a quotient is the quotient of the limits (as long as the denominator's limit isn't zero!).
$$ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M} $$ (provided M ≠ 0) - Power Rule: You can bring the limit "inside" a power or root.
$$ \lim_{x \to a} [f(x)]^n = L^n $$
You use these rules all the time without even thinking about it when you do Direct Substitution on a polynomial!
4. Limits at Infinity (Going on a Long Journey)
What happens to a function when `x` gets incredibly large? We call this a limit at infinity.
The most important idea here is this:
$$ \lim_{x \to \infty} \frac{1}{x} = 0 $$Think about it: 1/100, 1/1000, 1/1,000,000... As the denominator gets huge, the fraction gets closer and closer to 0. This works for any positive power of x: $$ \lim_{x \to \infty} \frac{c}{x^n} = 0 $$ for any constant `c` and `n > 0`.
How to Solve Limits of Rational Functions at Infinity
This is a required skill! There is a simple three-step process.
Example: Find $$ \lim_{x \to \infty} \frac{4x^2 - 3x + 5}{2x^2 + 7x - 1} $$
-
Find the highest power of `x` in the DENOMINATOR.
In our denominator `(2x² + 7x - 1)`, the highest power is `x²`. - Divide EVERY single term in the numerator AND the denominator by that power (`x²`). $$ \lim_{x \to \infty} \frac{\frac{4x^2}{x^2} - \frac{3x}{x^2} + \frac{5}{x^2}}{\frac{2x^2}{x^2} + \frac{7x}{x^2} - \frac{1}{x^2}} $$
- Simplify and evaluate the limit. $$ \lim_{x \to \infty} \frac{4 - \frac{3}{x} + \frac{5}{x^2}}{2 + \frac{7}{x} - \frac{1}{x^2}} $$ Now, as `x → ∞`, all the terms like $$ \frac{3}{x} $$, $$ \frac{5}{x^2} $$, etc., will go to 0! $$ \frac{4 - 0 + 0}{2 + 0 - 0} = \frac{4}{2} = 2 $$
So, the limit is 2.
Memory Aid: The Degree Shortcut
For rational functions $$ \frac{P(x)}{Q(x)} $$, compare the degree (highest power) of the numerator and denominator.
- If degree of top < degree of bottom, the limit is 0.
- If degree of top = degree of bottom, the limit is the ratio of the leading coefficients. (Like in our example, 4/2).
- If degree of top > degree of bottom, the limit is $$ \infty $$ or $$ -\infty $$ (the limit does not exist as a finite number).
Warning: While this shortcut is great for checking, you are often required to show the full working by dividing by the highest power of x in the denominator.
5. Two Very Special Limits You Must Know
These two limits are fundamental in calculus and are provided in your formula sheet, but you need to know how to use them. They often require a bit of manipulation.
The Special Trigonometric Limit
$$ \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 $$Important Note: This only works when $$ \theta $$ is in radians!
The trick to using this formula is to make the argument of `sin` match the denominator exactly.
Example: Find $$ \lim_{x \to 0} \frac{\sin(5x)}{2x} $$
We need the denominator to be `5x`, not `2x`. We can use algebra to make this happen!
$$ \lim_{x \to 0} \frac{\sin(5x)}{2x} = \lim_{x \to 0} \frac{\sin(5x)}{5x} \cdot \frac{5}{2} $$We multiplied by 5/5, which is just 1, and rearranged. Now we can use the limit rules:
$$ = \left( \lim_{x \to 0} \frac{\sin(5x)}{5x} \right) \cdot \frac{5}{2} $$As `x → 0`, `5x` also approaches 0. So the first part is our special limit, which equals 1.
$$ = (1) \cdot \frac{5}{2} = \frac{5}{2} $$The Special Exponential Limit
$$ \lim_{x \to 0} \frac{e^x - 1}{x} = 1 $$Similar to the `sin` limit, the goal is to make the exponent of `e` match the denominator.
Example: Find $$ \lim_{x \to 0} \frac{e^{3x} - 1}{x} $$
We need a `3x` in the denominator. Let's make it happen.
$$ \lim_{x \to 0} \frac{e^{3x} - 1}{x} = \lim_{x \to 0} \frac{e^{3x} - 1}{3x} \cdot 3 $$The first part is our special limit, which equals 1.
$$ = (1) \cdot 3 = 3 $$Key Takeaway: Section 5
When you see a limit as `x → 0` involving `sin(ax)` or `e^(ax) - 1`, your goal is to algebraically manipulate the expression so you can apply the two special limit formulas.
Chapter Summary: Your Limit-Finding Master Plan
Feeling overwhelmed? Don't be! Just follow this strategic process every time you see a limit problem.
-
What is `x` approaching?
- If `x → a` (a number): Go to Step 2.
- If `x → ∞`: Go to Step 3.
-
`x → a` Strategy:
- Try Direct Substitution. If it gives you a number, you're done!
- If you get $$ \frac{0}{0} $$, it's time for algebra.
- Factor and Cancel if you see polynomials.
- Rationalise with the Conjugate if you see square roots.
- If you see `sin(x)` or `e^x - 1` and `x → 0`, try to use the Special Limits.
-
`x → ∞` Strategy:
- This is almost always a rational function.
- Identify the highest power of `x` in the denominator.
- Divide every term by this power of `x`.
- Simplify and use the rule that $$ \frac{c}{x^n} \to 0 $$.
And that's it! Limits are a gateway to the rest of calculus. Practice these techniques, and you'll be in great shape. You've got this!