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

The word “limit” has several meanings in English Language but one of its meanings is a value.

In calculus, the word “limit” is used to represent a value of the function as the input approaches some value. In a function, an input is a variable.

Let’s learn the concept of the limit from a simple understandable example.

$2x+3$

It is a simple algebraic function and the variable $x$ represents the input of the function.

Let’s assume that we have to calculate the limit of the function as the variable $x$ closer to $5$.

We know that an operator for finding the limit of a function is simply denoted by $\lim$ and $x$ closer to $5$ is expressed as $x \,\to\, 5$ in calculus. Hence, the limit of the function $2x+3$ as $x$ approaches to $5$ is expressed in the following mathematical form.

$\displaystyle \large \lim_{x\,\to\,5}{\normalsize 2x+3}$

Remember that $x \,\to\, 5$ means we have to calculate the value of the function $2x+3$ by substituting a value, which is closer to $5$.

Firstly, find the difference between the numbers $4$ and $5$

$5-4 \,=\, 1$

There is no doubt about it that the number $4$ is near to $5$ but the difference between them is $1$, which indicates that there is some distance between them and it cannot be negligible. Hence, we cannot consider that the value of $4$ is closer to $5$.

Find the difference between the numbers $4.9$ and $5$.

$5-4.9 \,=\, 0.1$

$\implies$ $5-4.9 \,\approx\, 0$

In this case, the difference between $5$ and $4.9$ is $0.1$, which is approximately zero. Hence, we can take $4.9$ as a value closer to $5$. So, substitute $x \,=\, 4.9$ to find the limit of the function as $x$ approaches $5$.

$\displaystyle \large \lim_{x\,\to\,5}{\normalsize 2x+3}$ $\,=\,$ $2(4.9)+3$

$\implies$ $\displaystyle \large \lim_{x\,\to\,5}{\normalsize 2x+3}$ $\,=\,$ $9.8+3$

$\implies$ $\displaystyle \large \lim_{x\,\to\,5}{\normalsize 2x+3}$ $\,=\,$ $12.8$

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \large \lim_{x\,\to\,5}{\normalsize 2x+3}$ $\,\approx\,$ $13$

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

Once again, find the difference between $5$ and $4.99$.

$5-4.99 \,=\, 0.01$

$\implies$ $5-4.99 \,\approx\, 0$

The difference between $5$ and $4.99$ is $0.01$, which is also approximately zero. Therefore, we can take $4.99$ as a value closer to $5$. Hence, we can substitute $x = 4.99$ for evaluating the limit of the function as $x$ approaches $5$.

$\displaystyle \large \lim_{x\,\to\,5}{\normalsize 2x+3}$ $\,=\,$ $2(4.99)+3$

$\implies$ $\displaystyle \large \lim_{x\,\to\,5}{\normalsize 2x+3}$ $\,=\,$ $9.98+3$

$\implies$ $\displaystyle \large \lim_{x\,\to\,5}{\normalsize 2x+3}$ $\,=\,$ $12.98$

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \large \lim_{x\,\to\,5}{\normalsize 2x+3}$ $\,\approx\,$ $13$

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

Not only, $x = 4.9$ and $x = 4.99$. You can take any value which is closer to $5$ as input value to calculate the limit of the function $2x+3$. For example, $4.95$, $4.907$, $4.9999$ and so on.

Thus, we can calculate the limit of any function in the calculus.

Latest Math Topics

Latest Math Problems

Email subscription

Math Doubts is a free math tutor for helping students to learn mathematics online from basics to advanced scientific level for teachers to improve their teaching skill and for researchers to share their research projects.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.