The limit of the simple expression $4x$ has to calculate as $x$ approaches $7$ in this problem and this problem is useful for the beginners. The limit of the function $4x$ can be evaluated by substituting any value that is closer to $7$.

You can take any value that is closer to $7$, for example $x = 6.9917$. There are two reasons for this.

- The value of $6.9917$ is slightly less than $7$ but its approximate value is equal to $7$. In other words, $6.9917 \approx 7$.
- The difference between $7$ and $6.9917$ is also approximately small and negligible. In other words, $7-6.9917 = 0.0083$ and $0.0083 \approx 0$.

The above two points have cleared that the value of $6.9917$ closer to $7$ and substitute it to evaluate the limit of the function as $x$ approaches $7$.

$L \,=\, \displaystyle \large \lim_{x \,\to\, 7}{\normalsize (4x)}$

$\implies$ $L \,=\, 4(6.9917)$

$\implies$ $L \,=\, 4 \times 6.9917$

$\implies$ $L \,=\, 27.9668$

$\,\,\, \therefore \,\,\,\,\,\,$ $L \,\approx\, 28$

Therefore, the limit of the function $4x$ is equal to $28$ as $x$ approaches $7$.

It can also be obtained directly by substituting $x = 7$ in the given function.

$=\,\,\, 4(7)$

$=\,\,\, 4 \times 7$

$=\,\,\, 28$

Therefore, the limit of the function $4x$ as $x$ tends to $7$ is considered as the value of the function $4x$ at $x = 7$ in calculus.

$L \,=\, \displaystyle \large \lim_{x \,\to\, 7}{\normalsize (4x)}$

$\implies$ $L \,=\, 4(7)$

$\implies$ $L \,=\, 4 \times 7$

$\,\,\, \therefore \,\,\,\,\,\,$ $L \,=\, 28$

Theoretically, it is wrong to consider that the limit of the function $4x$ as $x$ approaches $7$ is equal to the value of the function at $x = 7$. However, it is acceptable to consider that they both are equal due to the negligible difference between the input values and also negligible difference between their corresponding values of the function.

Latest Math Topics

Latest Math Problems

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved