Limit of a Function

The word "limit" has often been heard in our daily lives. For example, someone says, "My patience limit is almost gone" or "The credit card you use is almost close to the limit." Words like boundaries, almost, and limits have meaning to or close to something, very close, but unable to reach or not can be exactly the same. The use of these words has a relationship with the word limit which we will study below.

Understanding the Concept of Limit Functions at a Point
Limit is the basic concept for differential calculus and integral calculus material. In mathematical language, limits explain the value of a function if it is approached from a certain point, Why should it be approached from a certain point and not exactly at that particular point? This is because not all functions have values ​​at all points. For example, the function $f (x) = \frac{x^2-1}{x-1}$ does not have a value (meaning) at x=1 because f(1) has an uncertain value of 0/0. If we take x values ​​from greater than 1 (from the right) and smaller than 1 (from the left) close to 1 then the value of f(x) tends to approach the value 2. This value 2 is called the limit value of the function $f (x) = \frac {x^2-1}{x-1}$ when x approaches 1 from the left and right. Then the limit value of the function is called the limit function $f (x) = \frac{x^2-1}{x-1}$. Mathematically written:

$\lim_{x \rightarrow 1} f (x) = \frac{x^2-1}{x-1} = 2$.

A function f(x) is said to have a limit value at point x=a if the limit function f(x) for x approaches a from the left is equal to the limit function f(x) for x close to a from the right. Suppose the limit function f(x) for x approaches a from the left is $L_1$ and the limit function f (x) for x approaches a from the right is $L_2$. If $L_1 \neq L_2$, the limit function f(x) for x approaches a no. Conversely, if $L_1 = L_2 = L$, the limit of the function f(x) for x approaches a is L. Mathematically, the definition of the limit of the function is given in the following definition.
Definition:
A function y = f (x) is defined for x around a (then a is called the limit point), then $\lim_{x \rightarrow a} f (x) = L$ if and only if $\lim_{x \rightarrow a^-} f (x) = \lim_{x \rightarrow a^+} f (x) = L$.
From the limit function definition above, the limit function search can be done using a numerical approach such as compiling a table of function values ​​by taking the function domain from the left and right of a limit point; and the function graph approach is to see the graphic image of the function both from the left and the right of the limit point to know intuition the limit value of the function exists or does not exist.

Able to apply the main limit theaters
The following are limit Theorems which are useful in determining the limits of a function. Because the function you want to specify the limit can be in the form of numbers, fillings, times, and for functions that have known limits. Because if we only use a numerical approach or this graph is very inefficient and effective.
1. If f(x)=k then $\lim_{x \rightarrow a} f (x) = k$ (for each constant k and a real number).
2. If k is a constant $\lim_{x \rightarrow a} k.f(x) = k \lim_{x \rightarrow a} f(x)$
3. $\lim_{x \rightarrow a} [f (x) + g (x)] = \lim_{x \rightarrow a} f (x) + \lim_{x \rightarrow a} g (x)$
4. $\lim_{x \rightarrow a} [f (x) - g (x)] = \lim_ {x \rightarrow a} f (x) - \lim_ {x \rightarrow a} g (x)$
5. $\lim_{x \rightarrow a} [f (x) \times g (x)] = \lim_{x \rightarrow a} f (x) \times \lim_{x \rightarrow a} g (x)$
6. $\lim_{x \rightarrow a} \frac{f (x)}{g (x)} = \frac {\lim_ {x \rightarrow a} f (x)}{\lim_{x \rightarrow a} g (x)}$
7. $\lim_{x \rightarrow a} [f (x)]^n = [\lim_{x \rightarrow a} f (x)]^n$
8. $\lim_{x \rightarrow a} {^n \sqrt{f (x)}} = {^n \sqrt{\lim_{x \rightarrow a} f (x)}}$
Determine the Limit of the Algebraic Function
Form $\lim_{x \rightarrow a} f (x)$
Determining $\lim_{x \rightarrow a} f (x)$ can be done by direct substitution method which is looking for the function value f(x) at x = a. Because, if f(x) has a meaningful value at x = a (defined at x=a) then $\lim_{x \rightarrow a} f (x) = f (a)$.

Form $\lim_{x \rightarrow \infty} \frac{f (x)}{g (x)}$
To specify $\lim_{x \rightarrow \infty} \frac{f (x)}{g (x)}$, we must first understand why $\lim_{x \rightarrow \infty} \frac{1}{x} = 0$. We understand it intuitively that if 1 is divided by many numbers into infinity, the result tends to go to 0. Note the graphical function image $f (x) = \frac{1}{x}$ following, when $x \rightarrow \infty$ then $f (x) \rightarrow 0$

Because $\lim_{x \rightarrow \infty} \frac{1}{x} = 0$ then for every n positive number and a real number, $\lim_{x \rightarrow \infty} \frac{a}{x^n} = 0$. With this knowledge, determining the limit of the algebraic function $\lim_{x \rightarrow \infty} \frac{f (x)}{g (x)}$ (certain form) is done by dividing the numerator f(x) and part denominator g(x) with $x^n$, where n is the highest power of f(x) or g(x).

Problems example:
$\lim_{x \rightarrow \infty} \frac{2x^2 + 3x-1}{x + 2}$
Settlement:
\begin{align} \lim_{x \rightarrow \infty} \frac{2x^2 + 3x-1}{x + 2} & = \lim_{x \rightarrow \infty} \frac{\frac{2x^2 + 3x-1}{x^2}}{\frac{x + 2}{x ^ 2}} \\ & = \lim_{x \rightarrow \infty} \frac{2+ \frac{3}{ x} - \frac{1}{x^2}}{\frac{1}{x} + \frac{2}{x ^ 2}} \\ & = \infty \end{align}

$\lim_{x \rightarrow \infty} \frac{(1-2x)^2}{\sqrt{8x^4-1}}$
Settlement:
\begin{align} \lim_{x \rightarrow \infty} \frac{(1-2x)^2}{\sqrt{8x^4-1}} & = \lim_{x \rightarrow \infty} \frac{\frac{(1-2x)^2}{x^2}}{\frac{\sqrt{8x^4-1}}{x^2}} \\ & = \lim_{x \rightarrow \infty } \frac{(\frac{1-2x}{x})^2}{\frac{\sqrt{8x ^ 4-1}}{\sqrt{x^4}}} \\ & = \lim_{ x \rightarrow \infty} \frac{(\frac{1}{x} - 2)^2}{\sqrt{8- \frac{1}{x^4}}} \\ & = \frac{4}{\sqrt{8}} \\ & = \frac{4}{2 \sqrt{2}} \\ & = \frac{2}{\sqrt{2}} \\ & = \sqrt{2} \end{align}

Form $\lim_{x \rightarrow \infty} [\sqrt{f (x)} - \sqrt{g (x)}]$
The limit of the function in the form of $\lim_{x \rightarrow \infty} [\sqrt{f (x)} - \sqrt{g (x)}]$ can be solved by multiplying by the opposite factor, $\frac{\sqrt{f (x)} + \sqrt{g (x)}}{\sqrt{f (x)} + \sqrt {g (x)}}$ so that it becomes the form $\lim_{x \rightarrow \infty} \frac{j (x)}{k (x)}$.
Problems example:
$\lim_{x \rightarrow \infty} [\sqrt{2x-1} - \sqrt{3x + 5}]$
Settlement:
\begin{align} & \lim_{x \rightarrow \infty} [\sqrt{2x-1} - \sqrt{3x + 5}] \\ & = \lim_{x \rightarrow \infty} [\sqrt{2x-1} - \sqrt{3x + 5}] \times \frac{\sqrt{2x-1} + \sqrt{3x + 5}}{\sqrt{2x-1} + \sqrt{3x + 5 }} \\ & = \lim_{x \rightarrow \infty} \frac{{2x-1} - (3x + 5)}{\sqrt{2x-1} + \sqrt{3x + 5}} \\ & = \lim_{x \rightarrow \infty} \frac{-x-6}{\sqrt{2x-1} + \sqrt{3x + 5}} = - \infty \end{align}

Form $\lim_{x \rightarrow \infty} [\sqrt{ax^2 + bx + c} - \sqrt{px^2 + qx + r} ]$
This form often appears in high school / equivalent national exam questions and can be completed quickly using the following conditions.
1. If a = p then $\lim_{x \rightarrow \infty} [\sqrt{ax^2 + bx + c} - \sqrt{px^2 + qx + r}] = \frac{bq}{2 \sqrt{a}}$
2. If a > p then $\lim_{x \rightarrow \infty} [\sqrt{ax^2 + bx + c} - \sqrt{px^2 + qx + r}] = \infty$
3. If a < p is $\lim_{x \rightarrow \infty} [\sqrt{ax^2 + bx + c} - \sqrt{px^2 + qx + r}] = - \infty$

Problems example:
Calculate $\lim_{x \rightarrow \infty} [\sqrt{3x^2-4x + 8} - \sqrt{3x^2-2x + 7}]$
Settlement:
Because a = p then
\begin{align} \lim_{x \rightarrow \infty} [\sqrt{3x^2-4x + 8} - \sqrt{3x^2-2x + 7}] & = \frac{bq}{2 \sqrt{a}} \\ & = \frac{-4 - (- 2)}{2 \sqrt{3}} \\ & = \frac{-2}{2 \sqrt{3}} \\ & = - \frac{1}{\sqrt{3}} \\ & = - \frac{\sqrt{3}}{3} \end{align}

Determining the Limit of Trigonometric Functions
In some cases, the limit resolution of trigonometric functions is almost the same as the completion of the limit of algebraic functions, for example by direct substitution methods. If the direct substitution method produces an indeterminate value then it is done by factoring method so that it is not indeterminate anymore. The trigonometric formulas you have learned and the main limit theorem are useful in completing the limits of trigonometric functions.
Problems example:
1. $\lim_{x \rightarrow \frac{\pi}{4}} sin \ x = sin \ \frac{\pi}{4} = \frac{1}{2} \sqrt{2}$
2. \begin{align} \lim_{x \rightarrow 0} (cos^2 \ x - sin^2 \ x) & = (\lim_{x \rightarrow 0} cos \ x)^2 - (\lim_{x \rightarrow 0} sin \ x)^2 \\ & = (1)^2- (0)^2 \\ & = 1 \end{align}
Because $sin \ 2x = 2sin \ x \ cos \ x$ then
\begin{align} \lim_{x \rightarrow 0} \frac{sin \ x}{sin \ 2x} & = \lim_{x \rightarrow 0} \frac{sin \ x}{2sin \ x \ cos \ x} \\ & = \lim_{x \rightarrow 0} \frac{1}{2 \ cos \ x} \\ & = \frac{1}{2 \ cos \ (0)} \\ & = \frac{1}{2 \times 1} \\ & = \frac{1}{2} \end{align}

The limits of the trigonometric function can also be solved using formulas. The trigonometric function formulas in question are:
$\lim_{x \rightarrow 0} \frac{sin \ x}{x} = \lim_{x \rightarrow 0} \frac{x}{sin \ x} = 1$
$\lim_{x \rightarrow 0} \frac{tan \ x}{x} = \lim_{x \rightarrow 0} \frac{x}{tan \ x} = 1$
The limit formula for the basic trigonometric function above can be expanded. Suppose u is a function of x and if $x \rightarrow 0$ then $u \rightarrow 0$, so that the formulas can be written to be:
1. $\lim_{u \rightarrow 0} \frac{sin \ u}{u} = \lim_{u \rightarrow 0} \frac{u}{sin \ u} = 1$
2. $\lim_{u \rightarrow 0} \frac{tan \ u}{u} = \lim_{u \rightarrow 0} \frac{u}{tan \ u} = 1$
Problems example:
$\lim_{u \rightarrow 0} \frac{1 - cos \ x}{x^2}$
Settlement:
If we do the direct substitution method, it turns out the result is in the form of indeterminate 0/0. This problem can be solved by using trigonometric similarities $cos \ 2x = 1 - 2 \ sin^2 \ x$ to change $1 - cos \ x$ so we can use the formula $\lim_{u \rightarrow 0} \frac{sin \ u}{u} = \lim_{u \rightarrow 0} \frac{u}{sin \ u} = 1$. Because $cos \ 2x = 1 - 2 \ sin^2 \ x$ then $cos \ x = 1 - 2 \ sin^2 \ (\frac{1}{2} x) \Leftrightarrow 1-cos \ x = 2 \ sin^2 \ (\frac{1}{2} x)$, and by specifying $u = \frac{1}{2} x$ then
\begin{align} \lim_{x \rightarrow 0} \frac{1 - cos \ x}{x^2} & = \lim_{x \rightarrow 0} \frac{2 \ sin^2 \ \frac{ 1}{2} x}{x^2} \\ & = \lim_{x \rightarrow 0} \frac{2 \ sin^2 \frac{1}{2} x}{x^2} \times \frac{\frac{1}{4}}{\frac{1}{4}} \\ & = \lim_{x \rightarrow 0} \frac{1}{2} \ \frac{sin^2 \ \frac{1}{2} x}{(\frac{1}{2} x)^2} \\ & = \lim_{u \rightarrow 0} \frac{1}{2} \ \frac{sin^2 \ u} {u^2} \\ & = \frac{1}{2} \lim_{u \rightarrow 0}(\frac{sin \ u}{u})^2 \\ & = \frac{1}{2} (\lim_{u \rightarrow 0} \frac{sin \ u}{u})^2 \\ & = \frac{1}{2}. (1)^2 \\ & = \frac{1}{2} \end{align}