Math Doubts

The Limit

The value of a function as its input approaches some value is called the limit.

Introduction

The word “Limit” has several meanings in English language and one of its meanings is a value. So, the limit is a value of a function as the input of the function approaches some value. In a function, the input is a variable.

Now, let us try to understand the concept of limit in calculus from an understandable simple example.

Example

Find the limit of the function $2x+3$ as $x$ approaches $5$.

Now, find the value of the given function by substituting $x$ is equal to $5$.

$=\,\,\, 2(5)+3$

$=\,\,\, 10+3$

$=\,\,\, 13$

The value of the function $2x+3$ is equal to $13$ at $x$ is equal to $5$.

Remember, it is wrong to consider the value of the function $13$ as the limit of the function because the value of the function is different to the limit of a function.

For calculating the limit of the function $2x+3$, we have to take a value, which should be very closer to $5$.

Case – 1

We obviously know that $4$ is a close value to the $5$. So, calculate the value of the function by substituting the $x$ is equal to $4$.

$=\,\,\, 2(4)+3$

$=\,\,\, 8+3$

$=\,\,\, 11$

But, the value $11$ is not the limit of the function because there is a reason for this. The number $4$ does not equal to $5$ but we have to calculate the value of the function as its variable closer to $5$.

Case – 2

Take $x = 4.9$. There is no doubt about it that $4.9$ is closer to $5$ than the number $4$.

$=\,\,\, 2(4.9)+3$

$=\,\,\, 9.8+3$

$=\,\,\, 12.8$

The value of the function is $12.8$ but it is also not considered as the limit of the function because the number $4.9$ never equals to $5$.

Case – 3

Take $x = 4.99$. In this case, the number $4.99$ is closer to $5$ than the numbers $4$ and $4.9$. Once again, evaluate the function by substituting $x$ is equal to $4.99$.

$=\,\,\, 2(4.99)+3$

$=\,\,\, 9.98+3$

$=\,\,\, 12.98$

The number $4.99$ does not equal to $5$ but its approximate value of $4.99$ is equal to $5$. In other words $4.99 \approx 5$. Therefore, the value of the function $12.98$ is called the limit.

However, $4.99$ is not only the number closer to $5$, there are infinite number closer to $5$. So, we can take any of them as the input for evaluating the limit.

For example $x = 4.9928$. Now, substitute it in the given function.

$=\,\,\, 2(4.9928)+3$

$=\,\,\, 9.9856+3$

$=\,\,\, 12.9856$

The value of $4.9928$ is not equal to $5$ but its approximate value is equal to $5$. In other words, $4.9928 \approx 5$. Therefore, the value of the function $12.9856$ is called the limit of the function.

In this case, we got two limits $12.98$ and $12.9856$ for $x = 4.99$ and $x = 4.9928$ respectively. If you take another closer value, you get another value as the limit for the given function but there should be only one limit for the given function. It confuses those who learn the limits newly but the concept of limit is very easy.

$4.99 \ne 4.9928$ but $4.99 \approx 5$ and $4.9928 \approx 5$. So, you can any value closer to $5$ to evaluate the limit of the function but the problem is we have two limits here. However, their approximate values are equal. $12.98 \ne 12.9856$ but $12.98 \approx 13$ and $12.9856 \approx 13$.

Therefore, the limit of the function $2x+3$ is $13$ as $x$ approaches $5$.

Case – 4

Take $x = 5.011$ and find the value of the function.

$=\,\,\, 2(5.011)+3$

$=\,\,\, 10.022+3$

$=\,\,\, 13.022$

In this case, the number $5.011$ does not equal to $5$ but its approximate value is equal to $5$. In other words, $5.011 \approx 5$. Therefore, the value of the function $13.022$ is also called the limit of the function. The approximate value of the number $13.022$ is also equal to $13$. In other words, $13.022 \approx 13$.

Therefore, the limit of the function $2x+3$ is also equal to $13$ as $x$ approaches $5$.

In calculus, the $x$ approaches $5$ means it is a number slightly either less than or greater than $5$ because their difference between them is approximately zero.

$5-4.99 = 0.01 \approx 0$ and also $5.011-5 = 0.011 \approx 0$.

In both cases, we take slightly either less or greater than the input value to find the value of the function and it is called the limit of the function.

The approximate values of the limits of both cases are equal to the value of the function $2x+3$ at $x = 5$. Hence, we take the limit of the function $2x+3$ as $x$ approaches $5$ in calculus as the value of the function $2x+3$ at $x = 5$ in calculus.

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