Limits of Functions

Limits in mathematics are the most fundamental concepts when it comes to understanding functions in calculus, where limits describe the behavior of a function as it approaches a specific point. Talking about the limit of a function is talking about the value that the function approaches when the input or variable moves closer to a certain number. This is one concept that is vital in terms of understanding derivatives, integrals, and continuity.

1.0What is a Limit?

A limit is a description of the value that a function approaches as the input (x) gets closer and closer to a certain point. This informs us of what is happening to the output, either y or f(x), as the input gets very close to some specific value, though need not reach it.

For example, if we have a function f(x), we can say:

${\lim }_{x\to a}f(x)=L$

This means that as 𝑥 approaches 𝑎, the function f(x) approaches the value L. In other words, a limit is the value of the function to which the function appears to be getting closer with our increasing closeness to a point.

2.0Types of Limits

There are different situations in which limits can occur. Some of the common types include:

1. Finite Limit: When the function approaches a finite value as 

x approaches a certain number. For example:

${\lim }_{x\to 2}(x+3)=5$

Here, as x it approaches 2, the function x+3 approaches 5.

2. Infinite Limit: When the function approaches infinity as 𝑥 approaches a certain value. This indicates that the function grows without bounds. For example:

${\lim }_{x\to 0}\frac{1}{x}=\infty$

As x gets closer to 0, the value

$\frac{1}{x}$ becomes larger and larger, approaching infinity.

3. Left-Hand and Right-Hand Limits: Sometimes, we consider the limit from one side only:

• Left-Hand Limit

$\left({\lim }_{x\to a-}\right)$ : The value of the function as x approaches a from the left(i.e., x is less than a). 

• Right-Hand Limit

$\left({\lim }_{x\to a+}\right)$ : The value of the function as x approaches a from the right (i.e., x is greater than a). 

If the left-hand and right-hand limits are equal, the overall limit exists at that point.

3.0Limits of Functions of Two Variables 

In many real-world problems, we have functions of two variables. These are functions depending on two different inputs that define the output, f(x,y). To obtain the limits of functions of two variables, we need to understand how the function changes as the inputs approach certain values. 

For example, in a function

$f(x,y)=\frac{x^2+y^2}{x+y}$

The limit of this function as x and y both approach 0 will depend on how we approach the point (0, 0) along different paths (for example, along the line x = y  or along the axis x = 0). The same can be written as: 

${\lim }_{(x,y)\to (a,b)}f(x,y)$

4.0How to Calculate Limits

There are different approaches to calculating limits, depending on the function and the situation. Here are some basic methods:

1. Direct Substitution: The simplest method is the direct substitution of the value x inside the function. If a function is continuous and does not feature any of the problems involved with division by zero, just substitute x=a .
   For example:

${\lim }_{x\to 3}\left(x^2+2x\right)=3^2+2(3)=9+6=15$

2. Factoring: If direct substitution leads to an indeterminate form such as

$\frac{0}{0}$ , you may have to factor the function first. After factoring, simplify the expressions inside the parentheses and substitute the value of x. 

 Example:

${\lim }_{x\to 2}\frac{x^2-4}{x-2}$

Factor the numerator:

$\frac{(x-2)(x+2)}{x-2}$

Now, cancel out x-2 and substitute x=2:

${\lim }_{x\to 2}(x+2)=2+2=4$

3. Rationalizing: If the function involves a square root, rationalizing the numerator or denominator may help in simplifying the limit.
4. Using L’Hopital's Rule: This rule can be applied when limits result in indeterminate forms

$\frac{0}{0},or\frac{\infty }{\infty }$ L’Hospital’s rule involves taking the derivative of the numerator and the denominator and then calculating the limit.

5.0Solved Problems on Limits of Functions

Problem 1: Evaluate

${\lim }_{x\to \frac{\pi }{2}}x/2Secx-tanx$

Solution: Put y = /2 – x. Then y0 and x/2. therefore

${\lim }_{x\to \frac{\pi }{2}}x/2Secx-tanx={\lim }_{y\to 0}\left[\sec \left(\frac{\pi }{2}-y\right)-\tan \left(\frac{\pi }{2}-y\right)\right]$

${\lim }_{y\to 0}\left(\mathrm{cosec}y-\cot y\right)$

${\lim }_{y\to 0}\frac{1}{\sin y}-\frac{\cos y}{\sin y}$

${\lim }_{y\to 0}\frac{1-\cos y}{\sin y}$

$(cos2y=1-2si{n}^{2}y;2si{n}^{2}y=1-cos2y;andsin2y=siny.cosy)$

$={\lim }_{y\to 0}\frac{\frac{2si{n}^{2}y}{2}}{2si{n}^{2}y/2.cosy/2}$

$={\lim }_{y\to 0}\tan y=0$

Problem 2: Evaluate

$={\lim }_{x\to 0}\frac{x-\sin x}{x^3}$

Solution: The given limit is the indeterminate form of 0/0, so we apply the L’hospital rule. 

Differentiate the numerator: 

$f^{\prime }(x)=\frac{d}{dx}(x-sinx)=1-cosx$

Differentiate the denominator: 

$g^{\prime }(x)=\frac{d}{dx}(x^3)=3x^2$

The limit becomes: 

${\lim }_{x\to 0}\frac{1-\cos \left(x\right)}{3x^2}$

By rechecking the limit and putting x — 0, the limit is still in indeterminate form, so we apply the L’hospital rule again. 

Differentiate the numerator again: 

$\frac{d}{dx}(1-cosx)=sinx$

Differentiate the denominator again: 

$\frac{d}{dx}3x^2=6x$

The limit becomes

${\lim }_{x\to 0}\frac{\sin x}{6x}$

We know that,

${\lim }_{x\to 0}\frac{\sin x}{x}=1$,Therefore, the limit becomes: 

$\frac{1}{6}{\lim }_{x\to 0}\frac{\sin x}{x}=\frac{1}{6}$

Problem3: Evaluate

${\lim }_{x\to 2}\frac{1}{x-2}-\frac{2\left(2x-3\right)}{x^{3-3x^2+2x}}$

Solution: we have

${\lim }_{x\to 2}\frac{1}{x-2}-\frac{2\left(2x-3\right)}{x^{3-3x^2+2x}}$

Taking the denominator of the limit

$x^3-3x^2+2x=x(x^2-3x+2)=x(x^2-2x-x+2)$

$=x[x(x-2)-1(x-2)]=x(x-2)(x-1)$

${\lim }_{x\to 2}\frac{1}{x-2}-\frac{2\left(2x-3\right)}{x\left(x-2\right)\left(x-1\right)}={\lim }_{x\to 2}\frac{x\left(x-1\right)-2{\left(2x-3\right)}^{}}{x\left(x-2\right)\left(x-1\right)}$

${\lim }_{x\to 2}\frac{x^2-5x+6}{x\left(x-2\right)\left(x-1\right)}={\lim }_{x\to 2}\frac{x^2-2x-3x+6}{x\left(x-2\right)\left(x-1\right)}$

${\lim }_{x\to 2}\frac{x\left(x-2\right)-3\left(x-2\right)}{x\left(x-2\right)\left(x-1\right)}={\lim }_{x\to 2}\frac{\left(x-2\right)\left(x-3\right)}{x\left(x-2\right)\left(x-1\right)}$

${\lim }_{x\to 2}\frac{x-3}{x\left(x-1\right)}=\frac{-1}{2}$