The best way to understand limits is to use intuition off-hand.
Let \(f(x)\) be a generic function of our choice.
The limit of \(f(x)\) as \(x\) approaches a certain number "a" is the value that \(f(x)\) is apperentely going to return when \(x\) becomes closer and closer to "a".
And it's denoted as what follows:
$$\lim_{x \to a} f(x)$$
$$Let f(x) = x + 2$$
The limit of \(f(x)\) as \(x\) approaches 1 is represented like this:
$$\lim_{x \to 1} f(x)$$
We know that \(f(x) = x + 2\), so we can subsstitute in the limit:
$$\lim_{x \to 1} x + 2$$
This is basically asking: "What is the value of the function when \(x\) iss almost equal to 1?"
In this case the answer is eassy, since when \(x\) is very close to \(1\), the whole function is almost equal to \(1 + 2\), which is \(3\).
Therefore, this is the easiest type of limit, where we can just substitute \(x\) by the number which the limit is tending to, like this:
$$\lim_{x \to 1} x + 2 = 1 + 2 = 3$$
Let's say this is the limit we're now trynna solve:
$$\lim_{x \to 2} \frac{x^2 -4}{x - 2}$$
Let's first test direct substitution:
$$\frac{x^2 - 4}{x - 2} = \frac{2^2 - 4}{2 - 2} = \frac{4 - 4}{2 - 2} = \frac{0}{0}$$
You see? Direct Substitution leads us to an indetermination!
Whenever the expression inside the limit develops into an indetermination, like \(\frac{0}{0}\), \(\frac{\infty}{\infty}\) or some other indetermination, you should always look for alternative ways to manipulate the expression to be able to solve the limit.
So, we are gonna first manipulate the expression a little bit:
Note that the numerator \(x^2 - 4\) can be written as \(x^2 - 2^2\)
OMG! A Remarkable Product! The Difference of Squares to be exact!
And the difference of squares says:
$$a^2 - b^2 = (a + b) \cdot (a - b)$$
So we have:
$$x^2 - 2^2 = (x + 2) \cdot (x - 2)$$
Put this product in the place of the old \(x^2 -4\) in the original expression:
$$\frac{x^2 - 2^2}{x - 2} = \frac{(x + 2) \cdot (x - 2)}{x - 2}$$
Note that we can cancel \(x - 2\) both in the numerator and in the denominator.
In ordinary cases, where \(x = 2\), we CANNOT cancel \(x - 2\), since that would bee ignoring the indetermination \(\frac{0}{0}\), which is a complete absurd!
However, we are treating limits, and \(x\) is NEVER equal to the tending number, it's just so close.
That's why we can cancel the \(x - 2\), since \(x\) is just very close to 2, but not exactly to 2.
Cancel \(x - 2\):
$$\frac{(x + 2) \cdot (x - 2)}{x - 2} = x + 2$$
And as we could reduce that whole fraction to a simpler function, we can now substitute in the original limit.
$$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} x + 2$$
We then reduced the whole "indetermination" to just a valid "direct substitution":
$$x + 2 = 2 + 2 = 4$$
Therefore:
$$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$$
Let's say we have this limit to solve:
$$\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x}$$
Let's first try Direct Substitution:
$$\frac{\sqrt{x + 1} - 1}{x} = \frac{\sqrt{0 + 1} - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}$$
Remember the tip from Example 1 , let's try to manipulate the expression.
Since the numerator is kinda complex to visualize, let's simplify it by introducing an Auxiliary Variable:
Let \(a = \sqrt{x + 1}\)
Then we have:
$$\frac{\sqrt{x + 1} - 1}{x} = \frac{a - 1}{x}$$
When there's a limit envolving roots, it's always recommended that you Rationalize it.
To rationalize it, you should multiply both the numerator and the denominator by the Conjugate.
To get the conjugate, just change the signal between the two terms. Look:
$$a - b \Rightarrow conjugate = a + b$$
$$a + b \Rightarrow conjugate = a - b$$
Rationalizing what we have:
$$\frac{a - 1}{x} = \frac{a - 1}{x} \cdot \frac{a + 1}{a + 1} = \frac{(a - 1) \cdot (a + 1)}{x \cdot (a + 1)}$$
We now have the Sum by the Difference, which can be simplified into the Difference of Squares, such as in Example 2
$$\frac{(a - 1) \cdot (a + 1)}{x \cdot (a + 1)} = \frac{a^2 - 1^2}{x \cdot (a + 1)}$$
Now it's better that we resume the substitution by the Auxiliary Variable like this:
$$a^2 = (\sqrt{x + 1})^2 = x + 1$$
$$\frac{a^2 - 1^2}{x \cdot (a + 1)} = \frac{x + 1 - 1}{x \cdot (\sqrt{x + 1} + 1)} = \frac{x}{x \cdot (\sqrt{x + 1} + 1)}$$
Cancel \(x\) in the numerator and the denominator:
$$\frac{1}{\sqrt{x + 1} + 1}$$
Just do Direct Substitution:
$$\frac{1}{\sqrt{0 + 1} + 1} = \frac{1}{\sqrt{1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$$
Therefore:
$$\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x} = \frac{1}{2}$$
Let \(f(x) = \frac{1}{x}\)
This is an "asymptotic function", that is: When \(x\) tends to \(\infty\), the whole function gets smaller and smaller, towards \(0\), and its value becomes almost \(0\).
Therefore:
$$x \to \infty \implies \frac{1}{x} \to 0$$
And it's also valid to when \(x\) tends to \(-\infty\), since it's the same case, but in stead of getting closer to \(0\) from positive numbers, it gets closer to \(0\) from negative numbers.
Therefore:
$$x \to -\infty \implies \frac{1}{x} \to 0$$
Let's solve:
$$\lim_{x \to \infty} \frac{2x + 1}{x}$$
When we try Direct Substitution, that leads us with \(\frac{\infty}{\infty}\), which is indeterminate.
So let's divide both the numerator and the denominator by \(x\), or just multiply them by \(\frac{1}{x}\):
$$\frac{2x + 1}{x} \cdot \frac{\frac{1}{x}}{\frac{1}{x}} = \frac{\frac{2x + 1}{x}}{\frac{x}{x}} = \frac{\frac{2x}{x} + \frac{1}{x}}{\frac{x}{x}}$$
Note that \(\frac{x}{x} = 1\) and \(\frac{2x}{x} = 2\).
Also keep in mind that our \(x\) is approaching \(\infty\), that means \(\frac{1}{x}\) will become \(0\) inside the limit.
Leaving us with the following:
$$\frac{\frac{2x}{x} + \frac{1}{x}}{\frac{x}{x}} = \frac{2 + 0}{1} = \frac{2}{1} = 2$$
Therefore:
$$\lim_{x \to \infty} \frac{2x + 1}{x} = 2$$
You can focus on the highest-degree terms when dealing with rational functions at infinity.
In that case, you could ignore the "\(+1\)" in the numerator, and just keep the "\(2x\)"".
In the denominator, you didn't have anything to do, since \(x\) was there by its own...
And then you could combine both ideas and get to \(\frac{2x}{x}\), which is \(2\).
Keep in mind that this only works when solving rational functions, that might not work with others!
One-sided limits are limits where \(x\) approaches a value from only one side.
For example, the limit as \(x\) approaches "a" FROM THE RIGHT of \(f(x)\) is this:
$$\lim_{x \to a^+} f(x)$$
And that means: \(x\) approaching value "a" from values greater than "a" itself.
And the limit as \(x\) approaches "a" FROM THE LEFT of \(f(x)\) is this:
$$\lim_{x \to a^-} f(x)$$
And that means: \(x\) approaching the value "a" from values less than "a" itself.
A two-sided limit exists only if the left-hand limit and the right-hand limit are equal.
Let's say we have:
$$\lim_{x \to 1} x^2$$
The left-hand limit is:
$$\lim_{x \to 1^-} x^2$$
And when \(x\) becomes closer to \(1\) from the left, the function becomes closer to \(1^2\), which is \(1\):
$$\lim_{x \to 1^-} x^2 = 1$$
The right-hand limit is:
$$\lim_{x \to 1^+} x^2$$
And when \(x\) becomes closer to \(1\) from the right, the function becomes closer to \(1^2\) as well, which is \(1\):
$$\lim_{x \to 1^+} x^2 = 1$$
Therefore, we can see that both one-sided limits are equal to each other. That means the original limit exists and it's also equal to them:
$$\lim_{x \to 1^-} x^2 = \lim_{x \to 1^+} x^2 = \lim_{x \to 1} x^2 = 1$$
Let's say we have:
$$\lim_{x \to 0} \frac{1}{x}$$
The left-hand limit is:
$$\lim_{x \to 0^-} \frac{1}{x}$$
And when \(x\) becomes closer to \(0\) from the left, the function becomes closer to \(\frac{1}{\text{very small negative number}}\), which is close to \(-\frac{1}{\text{very small number}}\), which tends to -\(\infty\):
$$\lim_{x \to 0^-} \frac{1}{x} = -\infty$$
The right-hand limit is:
$$\lim_{x \to 0^+} \frac{1}{x}$$
And when \(x\) becomes closer to \(0\) from the right, the function becomes closer to \(\frac{1}{\text{very small number}}\), which tends to \(\infty\):
$$\lim_{x \to 0^+} \frac{1}{x} = \infty$$
Therefore, we can see that both one-sided limits are different from each other. That means the original limit DOES NOT exist!
$$\lim_{x \to 0} \frac{1}{x} \text{ does not exist}$$
A function is continuous at a point if it satisfies the 3 conditions below:
1. \(f(a)\) must exist
2. \(\lim_{x \to a} f(x)\) must exist
3. Both of those values must be equal
"a" is simply any chosen point in the function.
Let's reuse previous examples:
Let \(f(x) = x^2\)
Note that for any point "a" chosen, it's true what follows:
\(f(a) = a^2\), it exists \(\checkmark\)
\(\lim_{x \to a} f(x) = a^2\), it exists \(\checkmark\)
And note that \(f(a)\) and \(\lim_{x \to a} f(x)\) are equal \(\checkmark\)
That means \(f(x) = x^2\) is a continuous function.
Let \(f(x) = \frac{1}{x}\)
When we choose \(a = 0\) as our point, look what happens:
\(f(a) = \frac{1}{0}\) it's undefined \(\implies\) does not exist \(\times\)
As we saw in the One-Sided Limits topic, the function \(f(x) = \frac{1}{x}\) has different one-sided limits on the point \(a = 0\). They are:
$$\lim_{x \to 0^-} \frac{1}{x} = -\infty$$
$$\lim_{x \to 0^+} \frac{1}{x} = \infty$$
So the limit does not exist \(\times\)
Therefore we can claim \(f(x) = \frac{1}{x}\) is not a continuous function, since when we choose \(a = 0\), the conditions are not met.
Yeah! You can make operations with limits! And it's pretty simple:
$$\lim {[f(x) + g(x)]} = {[\lim {f(x)}]} + {[\lim {g(x)}]}$$
$$\lim {[f(x) - g(x)]} = {[\lim {f(x)}]} - {[\lim {g(x)}]}$$
$$\lim {[f(x) \cdot g(x)]} = {[\lim {f(x)}]} \cdot {[\lim {g(x)}]}$$
$$\lim \frac{f(x)}{g(x)} = \frac{\lim{f(x)}}{\lim {g(x)}}$$
$$g(x) \neq 0$$
Let \(f(x) = x^2 \cdot \sin\left(\frac{1}{x}\right)\)
Let's find \(\lim_{x \to 0} f(x)\)
We know that the sine of anything is always inside the interval from -1 to 1, that is:
$$-1 \le \sin\left(\frac{1}{x}\right) \le 1$$
You can multiply everything by \(x^2\) to make \(f(x)\) appear:
$$-x^2 \le x^2 \cdot \sin\left(\frac{1}{x}\right) \le x^2$$
Now that we've expressed \(f(x)\) inside this chain, we can apply Squeeze Theorem:
Whenever you have a chain of functions,
$$f(x) \le g(x) \le h(x)$$
And the limit of the outter functions are the same,
$$\lim f(x) = \lim h(x) = L$$
The limit of the inner function is also the same.
$$\lim g(x) = L$$
And you can use this theorem to solve \(\sin\), \(\cos\) or other functions that are bounded by an interval.
Applying Squeeze Theorem we have:
$$\lim_{x \to 0} -x^2 = 0$$
$$\lim_{x \to 0} x^2 = 0$$
The outter ones are equal \(\checkmark\)
Therefore we found the limit we were trynna solve:
$$\lim_{x \to 0} \left[x^2 \cdot \sin\left(\frac{1}{x}\right)\right] = 0$$
There aree some famous/classic limits that you'd really like to know.
However, they'd be kinda hard to explain, since you'd have to know Derivatives.
So I'm just gonna tell you the tricks, but the deep reassons only when we get to Derivatives.
Let's start with this:
$$\lim_{\theta \to 0} \sin(\theta)$$
When \(\theta \to 0\), \(\sin(\theta) \to \theta\), so:
$$\lim_{\theta \to 0} \sin(\theta) = \theta$$
Take a look at this one:
$$\lim_{\theta \to 0} \cos(\theta)$$
When \(\theta \to 0\), \(\cos(\theta) \to 1\), so:
$$\lim_{\theta \to 0} \cos(\theta) = 1$$
The last one is this:
$$e = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n$$
\(e\) is a very special number. It's the base of the natural logarithm: \(\ln(x)\)