**Understanding limits**

Limits are vital in understanding the behavior of functions as they approach certain values, be it a specific number or infinity. To truly grasp calculus, it is imperative to comprehend the nuances of limits. In this article, we will delve deep into calculating limits calculus, from their definitions to different types with examples and solutions.

Understanding limits is crucial for various calculus concepts, such as continuity, derivatives, and integrals, and it plays a central role in solving problems related to rates of change, motion, and many other applications in mathematics and science.

## Definition of a limit calculus

In calculus, the concept of a limit is a fundamental idea that describes the behavior of a function as it gets closer and closer to a specific value or as it approaches infinity or negative infinity. The formal definition of a limit in calculus is as follows.

Let *(**x*** )** be a function defined for values of “ near a point “ , except possibly at “ itself. We say that the limit of

*(*

*x***as “ approaches “ is**

*)***“**“, denoted as

$$ \lim_{x\rightarrow c}\;\;\;\;f\;(x)\;\;=\;L $$

If for every positive number *ϵ*, there exists a positive number *δ* such that, whenever 0<∣*x*−*c*∣<*δ*, it follows that ∣*f*(*x*)−*L*∣<*ϵ*

In simpler terms, this definition means that as *x* gets closer to the value *c*, the function *f*(*x*) gets closer and closer to the limit *L*. The formal definition uses the concept of “epsilon (*ϵ*) and delta (*δ*)” to describe this idea of closeness and ensures that *f*(*x*) remains within a certain range (*ϵ*) of the limit (*L*) when *x* is within a certain range (*δ*) of the point *c*.

### How to calculate limits in calculus

To calculate a limit, follow these steps:

1) Substitute the value $c$ into the function $f(x)$.

2) If direct substitution yields a defined value, that value is the limit.

3) If direct substitution results in an indeterminate form (such as 0/0 or ∞/∞), use techniques like factoring, simplification, or L’Hôpital’s Rule to resolve the limit.

Also we can use online calculator to find the limit

**Types of Limits with examples**

**One-Sided Limits:**

One-sided limits focus on the behavior of a function as it approaches a point from either the left side or the right side. These limits are particularly helpful when you’re dealing with functions that exhibit different behaviors from each side of a point.

**Left-Sided Limit (Lim(x→a-) f(x))**: This describes how a function behaves as it approaches ‘a’ from the left side (values less than ‘a’). It’s denoted as Lim(x→a-) f(x).

**Right-Sided Limit (Lim(x→a+) f(x))**: This focuses on the behavior as a function approaches ‘a’ from the right side (values greater than ‘a’). It’s denoted as Lim(x→a+) f(x).

**Example -1 **

Consider the function *f*(*x*)=∣*x*∣. For the left-sided limit as *x* approaches 0, Lim(x→0-) *f*(*x*) is 0 because the function approaches 0 from the left side. For the right-sided limit as *x* approaches 0, Lim(x→0+) *f*(*x*) is also 0.

**Example- 2**

Consider the function \(f(x) = \begin{cases} x + 2 & \text{if } x < 0 \\ 3x & \text{if } x \geq 0 \end{cases}\)

For the left-sided limit as \(x\) approaches 0, Lim(x→0-) \(f(x)\) is \(2\) because for values less than 0, the function behaves like \(x + 2\) and approaches 2 as \(x\) gets close to 0 from the left side.

For the right-sided limit as \(x\) approaches 0, Lim(x→0+) \(f(x)\) is \(0\) because for values greater than or equal to 0, the function behaves like \(3x\) and approaches 0 as \(x\) gets close to 0 from the right side.

**Limits at a Point:**

These limits concentrate on how a function behaves near a specific point ‘a’. A limit at a point exists if the function approaches a specific value ‘L’ as ‘x’ gets arbitrarily close to ‘a’. It’s denoted as Lim(x→a) f(x) = L.

**Example-1**:

For the function *f*(*x*)=*x*^{2}, the limit as *x* approaches 2, Lim(x→2) *f*(*x*), is 4, because as *x* gets close to 2, the function approaches 4.

**Example-2:**

Function: \(f(x) = x^3 – 2x^2 + 3x – 1\)

Explanation: Finding the limit as \(x\) approaches 2.

Result: \( \lim_{{x \to 2}} f(x) = 5 \)

**Example -3:**

Function: \(f(x) = \frac{\sin(x)}{x}\)

Explanation: Calculating the limit as \(x\) approaches 0.

Result: \( \lim_{{x \to 0}} \frac{\sin(x)}{x} = 1 \) (Using L’Hôpital’s Rule)

**Note**: L’Hôpital’s Rule states that if you have an indeterminate form (0/0 or ∞/∞), you can take the derivative of the numerator and denominator and then try to find the limit again.

**Limits at Infinity:**

Limits at infinity help understand how functions behave as the input approaches positive infinity or negative infinity. These limits are denoted as Lim(x→∞) f(x) or Lim(x→-∞) f(x).

**Example-1**: Limit at Infinity (Lim(x→∞) f(x))

Function: \(f(x) = \frac{1}{x}\)

Explanation: We are determining the limit as \(x\) approaches positive infinity.

Result: \(\lim_{x \rightarrow \infty} \frac{1}{x} = 0\)

**Example-2:** Limit at Negative Infinity (Lim(x→-∞) f(x))

Function: \(f(x) = e^x\)

Explanation: We are exploring the limit as \(x\) approaches negative infinity.

Result: \(\lim_{{x \to -\infty}} e^x = 0\)

**Infinite Limits:**

Some functions exhibit unbounded growth or decrease as *x* approaches a specific value. Infinite limits can be either positive or negative infinity and are denoted as Lim(x→a) f(x) = +∞ or Lim(x→a) f(x) = -∞.

**Example-1: **Infinite Limit (Lim(x→a) f(x))

Function: \(f(x) = \frac{1}{x}\)

Explanation: We’re analyzing the limit as \(x\) approaches 0.

Result: \(\lim_{{x \to 0}} \frac{1}{x} = +\infty\).

**Example-2**: Infinite Limit (Lim(x→a) f(x))

Function: \(f(x) = -\frac{1}{x^2}\)

Explanation: We’re investigating the limit as \(x\) gets close to 0.

Result: \(\lim_{{x \to 0}} -\frac{1}{x^2} = -\infty\).

**Discontinuous Limits:**

Discontinuous limits occur when a function has a discontinuity at a particular location. In these cases, the left-sided limit and the right-sided limit do not coincide, and the limit may not exist.

**Example**-1: Discontinuous Limit (x = 0)

Function: \(f(x) = \frac{1}{x}\)

Explanation: We are examining the left-sided and right-sided limits at \(x = 0\).

Result: The limit at \(x = 0\) does not exist due to non-matching left-sided and right-sided limits.

**Example-2: **Discontinuous Limit (x = 1)

Function: \(f(x) = \begin{cases} 1 & \text{if } x < 1 \\ 2 & \text{if } x \geq 1 \end{cases}\)

Explanation: We are studying the left-sided and right-sided limits at \(x = 1\).

Result: The limit at \(x = 1\) does not exist because of differing left-sided and right-sided limits.

## Properties of limits with explanations:

** Constant Multiple Rule:**

You can factor a constant that is multiplicative out of a limit.

If \(c\) is a constant, then \(\lim_{{x \to a}} c \cdot f(x) = c \cdot \lim_{{x \to a}} f(x)\).

**Sum/Difference Rule:**

To consider the limit of a sum or difference, you can select the limits individually and put them back with the corresponding sign. This rule works regardless of the number of functions separated by “**+**” or “**–**“.

**Explanation:** If you have a limit of a sum or difference of two functions, like **$Lim(x→a)[f(x)+g(x)]$,** you can find the limit of each function separately and then add or subtract their limits.

\(\lim_{{x \to a}} [f(x) \pm g(x)] = \lim_{{x \to a}} f(x) \pm \lim_{{x \to a}} g(x)\)

**Product Rule:**

Consider the limit of a product in a similar way to the limit of sums or differences. Just select the limit of the factors and apply it. This rule is not limited to only two functions.

**Explanation:** If you have a limit of a product of two functions, such as **$Lim(x→a)[f(x)⋅g(x)]$**, you can find the limits of both functions individually and then multiply their limits.

\(\lim_{{x \to a}} [f(x) \cdot g(x)] = \lim_{{x \to a}} f(x) \cdot \lim_{{x \to a}} g(x)\).

**Quotient Rule:**

For the limit of a quotient, you only need to be concerned if the limit of the denominator is zero. Division by zero leads to an error, so the limit of the denominator must not be zero.

When finding the limit of a quotient, like **$Lim(x→a) [g(x)/f(x) ]$ **you can’t divide by zero. Therefore, you need to make sure that the limit of the denominator,** $g(x)$**, is not zero at the point you are approaching, **$a$**.

\(\lim_{{x \to a}} \frac{f(x)}{g(x)} = \frac{\lim_{{x \to a}} f(x)}{\lim_{{x \to a}} g(x)}\) (provided \(\lim_{{x \to a}} g(x) \neq 0\)).

**Limit of a Constant:**

The limit of a constant is only a constant. You can easily understand this by looking at the graph of a constant function, such as **$f(x)=c$**.

**Explanation:** If you take the limit of a constant value, like **$Lim(x→a)c$**, the limit is just that constant value, **$c$.** This is because the value of the function doesn’t change as **$x$** approaches **$a$.**

\(\lim_{{x \to a}} c = c\) (where \(c\) is a constant).

## Limits examples with solutions

**Example 1:** Simple Linear Function:

Find \(\lim_{{x \to 3}} (2x – 1)\).

Solution: This is a simple linear function. To calculate the limit, plug in $x=3$ into the function

\(\lim_{{x \to 3}} (2x – 1) = 2(3) – 1 = 5\).

The limit is 5.

**Example 2:** Quadratic Function

Find \(\lim_{{x \to -2}} (x^2 + 3x + 2)\).

Solution: For a quadratic function, simply substitute the value of

\(\lim_{{x \to -2}} (x^2 + 3x + 2) = (-2)^2 + 3(-2) + 2 = 0\).

The limit is 0.

**Example 3**: Rational Function

Find \(\lim_{{x \to 2}} \left(\frac{x^2 – 1}{x – 1}\right)\).

Solution: This is a rational function. Try direct substitution

\(\lim_{{x \to 2}} \left(\frac{x^2 – 1}{x – 1}\right) = \frac{2^2 – 1}{2 – 1} = 3\).

The limit is 3.

**Example 4:** Given function: \(f(x) = \frac{x^2 – 4}{x – 2}\)

Find the limit as \(x\) approaches 2, i.e., \(\lim_{{x\to 2}} f(x)\).

Solution:

Direct Substitution leads to an indeterminate form (\(0/0\)).

Factoring: Factor the numerator to get \(\frac{(x+2)(x-2)}{x-2}\).

Cancel out the common factor of \(x-2\).

\(\frac{x+2}{1} = x+2\)

Substitute \(x = 2\) into the simplified expression.

\(f(2) = 2 + 2 = 4\)

**Example 5:**

Given function : \[f(x) = \frac{x^2 – 4}{x^2 – 5x + 6}\]

Find the limit as \(x\) approaches 2, i.e., \(Lim(x→2) f(x)\).

Solution:

Direct Substitution leads to an indeterminate form (\(\frac{0}{0}\)).

Factor both the numerator and denominator:

\[ \frac{(x+2)(x-2)}{(x-2)(x-3)} \]

Cancel out the common factor of \((x-2)\):

\[ \frac{x+2}{x-3} \]

Substitute \(x = 2\) into the simplified expression:

\[ f(2) = \frac{2+2}{2-3} = -4 \]

\[ Lim(x→2) f(x) = -4 \]

**Example 6:**

Given function: \(f(x) = \frac{\sin(2x)}{x}\)

Find the limit as \(x\) approaches 0, i.e., \(Lim(x→0) f(x)\).

Solution: Apply L’Hôpital’s Rule:

Take the derivative of the numerator: \( \frac{d}{dx}[\sin(2x)] = 2\cos(2x) \).

Take the derivative of the denominator: \( \frac{d}{dx}[x] = 1 \).

The limit becomes \( Lim(x→0) \frac{2\cos(2x)}{1} \).

Substitute \(x = 0\) into the new expression:

\( Lim(x→0) \frac{2\cos(2(0))}{1} = 2\cos(0) = 2 \).

**Example 7:**

Given function: \(f(x) = \frac{e^x – 1}{x}\)

Find the limit as \(x\) approaches 0, i.e., \(Lim(x→0) f(x)\).

Solution:

Direct Substitution leads to an indeterminate form (\(\frac{0}{0}\)).

Apply L’Hôpital’s Rule:

Take the derivative of the numerator: \( \frac{d}{dx}[e^x – 1] = e^x \).

Take the derivative of the denominator: \( \frac{d}{dx}[x] = 1 \).

The limit becomes \( Lim(x→0) \frac{e^x}{1} \).

Substitute \(x = 0\) into the new expression:

\( Lim(x→0) \frac{e^0}{1} = 1 \).

**Summary :**

Limits in calculus are a fundamental concept that provides insights into how functions behave as they approach specific values. They play a pivotal role in various mathematical and scientific applications. In this comprehensive article, we’ve covered the definition of a limit in calculus, different types and properties of limits. In addition to explaining these concepts, we’ve provided numerous examples with step-by-step solutions of limit calculations and also covered calculating limits involves direct substitution or techniques like factoring and L’Hôpital’s Rule when faced with indeterminate forms.

Thank you for reading this article on* limits in calculus*. We hope you found it an informative and valuable article. If you have any comments, questions, or suggestions, please feel free to leave them below.

**Related Articles:**

Integral and Double Integral calculus Example with Solution

Two’s complement steps | turn decimal numbers or binary number into 2s complement

Time and Work Aptitude | Formulas with Questions and Answers

Coordinate geometry introduction | cartesian plane, X & Y-coordinate

Frustum of cone formulas with examples | Surface area and Volume

Surface Area and Volume of a Prism Formulas, rectangular and triangular

Formulas for Right circular cylinder, Oblique Cylinder & Hollow Cylinder