For solving functions in limits, there are also complex functions or combinations of two or more functions that may be used. So, for evaluating such types of limits we will use some simple rules of limits and also some important limit theorems here:
So before the learning of general rules of limits, let's take an overview of limit theorems.
Sandwich Theorem:
The sandwich theorem can be observed by its name like a sandwich has three corners. Similarly, we will see a relation or theorem between three functions i.e f(x),g(x) and h(x).
The main idea behind the sandwich theorem which is also known as the “Squeeze Theorem” is that if we have three functions like f(x),g(x), and h(x) in such a way that
$$ g(x) \;\le\; f(x) \;\le\; h(x) $$
So, f(x) is between g(x) and h(x)
Now, for a limit approaching some number like:
$$ \lim\limits_{x \to a} g(x) = \lim\limits_{x \to a} h(x) = L $$
Then, the limit for middle function f(x) will also the same L:
$$ \lim\limits_{x \to a} f(x) = L $$
Conclusion:
So, we concluded from sandwich theorem that if the small function g(x) and the large function h(x) has a value of L then the one in the middle f(x) should have the same limit L. this is the main idea behind the squeeze theorem but let's work on some examples to apply squeeze theorem.
Squeeze Theorem Examples:
Let's say that:
f(x) is in between two function
$$ 8-x^3 \;\le\; f(x) \;\le\; 8+x^3 $$
Now, what’s the answer for
$$ \lim\limits_{x \to 0} f(x) = ? $$
we figure this out well let's apply the limit to every expression
$$ \lim\limits_{x \to 0} (8-x^3) \;\le\; \lim\limits_{x \to 0} f(x) \;\le\; \lim\limits_{x \to 0}(8+x^3) $$
By using a simple substitution
$$ 8-(0)^3 \;\le\; \lim\limits_{x \to 0} f(x) \;\le\; 8+(0)^3 $$ $$ 8 \;\le\; \lim\limits_{x \to 0} f(x) \;\le\; 8 $$
So, by using the squeeze theorem we can say
If
$$ \lim\limits_{x \to 0} (8-x^3) = \lim\limits_{x \to 0} (8+x^3) = 8 $$
Then
$$ \lim\limits_{x \to 0} f(x) = 8 $$
L hôpital's Rule:
Let's say if you want to find the limit as:
$$ \lim\limits_{x \to a} \frac{f(x)}{g(x)}=? $$
If f(x) divided by g(x) have in indeterminate function form like
$$ \frac{0}{0} or \frac{\infty}{\infty}$$
These are indeterminate forms because we do not know what exactly is 0/0 or Infinity/infinity.
Then,
$$ \lim\limits_{x \to a} \frac{f(x)}{g(x)} = \lim\limits_{x \to a} \frac{f^{'}(x)}{g^{'}(x)} \;\;\;\;\;\;\;\;\;\; \forall f(x) \; \text{&} \; g(x) = 0 \;or\; \infty $$
Which is basically a L hôpital's Rule for solving limits at the indeterminate form
Conclusion:
The basic idea of L hôpital's rule if you can't find the limit in a simple form, take the derivative of the numerator and the denominator separately and you might be able to find the derivative in this form.
$$ \lim\limits_{x \to a} \frac{f(x)}{g(x)} = \lim\limits_{x \to a} \frac{f^{'}(x)}{g^{'}(x)} \;\;\;\;\;\;\;\;\;\; \forall f(x) \; \text{&} \; g(x) = 0 \;or\; \infty $$
Now make a sense of clear understanding of L hôpital's rule by practical examples.
Lopital Rule Example:
What is a limit as:
$$ \lim\limits_{x \to 3} \frac{x^2-9}{x-3}=? $$
By simple substitution, we have an indeterminate form here
$$ \lim\limits_{x \to 3} \frac{x^2-9}{x-3} = \frac{(3)^2-9}{3-3} = \frac{0}{0} $$
Let’s take the derivative of numerator and denominator according to L hôpital's rule.
$$ \lim\limits_{x \to 3} \frac{dy}{dx} \left( \frac{x^2-9}{x-3} \right) = \lim\limits_{x \to 3} \frac{2x}{1}$$
Now, substitute x by 3
$$ \lim\limits_{x \to 3} \frac{dy}{dx} \left( \frac{x^2-9}{x-3} \right) = 2(3)$$ $$ \lim\limits_{x \to 3} \frac{dy}{dx} \left( \frac{x^2-9}{x-3} \right) = 6 $$
In this way, by using lopital rule you can come out from indeterminate form for solving limits.
You can also confirm your limit function value by the factoring method.
General Rules of Limits
We are using several limit laws/rules for evaluating limits having arithmetic combinations of functions. We will see some basic limit laws and general rules of limits which will be helpful in solving limits:
Consider two functions:
$$ \lim\limits_{x \to a}f(x)=L \;\; \text{&} \;\; \lim\limits_{x \to a}g(x)=M $$
Here, L and M are values of functions at point “a”.
So rules for these functions are
- Sum Rule:
- $$\lim\limits_{x \to a} \left( f(x)+g(x) \right) = L+M $$
- Difference Rule:
- $$\lim\limits_{x \to a}\left( f(x)-g(x) \right)= L-M $$
- Constant Multiple Rule:
- $$\lim\limits_{x \to a}\left( k*f(x) \right)= k*L $$
Where k = constant (Real Number).
- Limit Product Rule:
- $$\lim\limits_{x \to a}\left( f(x)*g(x) \right)= L*M $$
- Quotient Rule:
- $$\lim\limits_{x \to a} \frac{f(x)}{g(x)}= \frac{L}{M} $$
Where M can never be 0.
- Limit Power Rule:
- $$ \lim\limits_{x \to a} \left[ f(x)^n \right] = L^n $$
Where n = positive integer.
- Root Rule:
- $$ \lim\limits_{x \to a} \sqrt[n]f(x) = \sqrt[n]L= L^\frac{1}{n} $$
Where n = positive integer.