Roughly speaking, finding the limit involves examining the behaviour of a function \(f(x)\) as \(x\) approaches a real number \(a\) that may or may not be in the domain of \(f\), that is,
\[\lim_{x\to a}f(x)=L\text{, where }L\text{ is a real number},\]
then it is said that the limit exists. Otherwise, the limit does not exist.
It is important to remember that limits describe the behaviour of a function near a particular point, but not necessarily at the point itself!
Computation of limits in some cases
Case 1. Direct substitution rule
Example.
\(\mathrm{Find~}\lim_{x\to-2}\frac{x^3+2x^2-1}{5-3x}.\)
Solution.
By substituting \(x\)=-2,
\(\lim_{x\to-2}\frac{x^3+2x^2-1}{5-3x}=\frac{(-2)^3+2(-2)^2-1}{5-3(-2)}=-\frac{1}{11}\)
Case 2.
If the fraction is of the form \(0/0\) and \(\frac\infty\infty,\), then, sometimes, there is HOPE!
Try to manipulate \(f(x)\) by rationalizing, factoring, etc. in order to cancel. Alternatively, L'Hôpital's Rule can be used
Example
Compute \(\lim_{x\to2}\frac{x-4}{\sqrt{x}-2}.\)
Solution
The fraction has the form \(0/0\). HOPE!
By using the difference of squares,
\(\begin{aligned}
\lim_{x\to4}\frac{x-4}{\sqrt{x}-2}& =\lim_{x\to4}\frac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x}-\sqrt{2}} \\
&=\lim_{x\to4}\sqrt x+\sqrt2=\sqrt4+\sqrt2 \\
&=2+\sqrt{2}
\end{aligned}\)
Case 3.
If the fraction is of the form \(1/0\), then the limit does not exist. (Note that 1 in the numerator can be any non-zero number)
Example
Evaluate \(\lim_{x\to-3}\frac{x^2-x+12}{x+3},\), if possible.
Note that \(f(x)\) cannot be reduced.
The limit of the numerator is \(\lim_{x\to-3}(x^2-x+12)=24,\) which is not equal to zero.
The limit of the denominator is \(\lim_{x\to-3}x+3=0.\)
Thus, the limit does not exist.
Squeeze Theorem
Suppose that \( g(x) \leq f(x) \leq h(x) \) for all \( x \) in an open interval containing \( a \), except possibly at \( x = a \) itself and that \( \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L \). Then, \( \lim_{x \to a} f(x) = L \).
Example.
Evaluate \( \lim_{x\to0} x^2 e^{\sin(1/x)} \).
Solution.
Using the fact that \( -1 \leq \sin(x) \leq 1 \) in order to create an inequality,
\[
-1 \leq \sin(1/x) \leq 1
\]
\[
e^{-1} \leq e^{\sin(1/x)} \leq e^1
\]
\[
x^2 e^{-1} \leq x^2 e^{\sin(1/x)} \leq x^2 e
\]
Since \( x^2 e^{-1} \leq x^2 e^{\sin(1/x)} \leq x^2 e \) and \( \lim_{x\to0} x^2 e^{-1} = \lim_{x\to0} x^2 e = 0 \), \(
\lim_{x\to0} x^2 e^{\sin(1/x)} = 0\) by the Squeeze Theorem. (Note that the limit cannot be evaluated by direct substitution as the limit of the exponential function, \( e^{\sin(1/x)} \), does not exist.)