Question

How do you find the limit of $2x + 5$ as $x \to {7^ - }$?

Answer 1

To solve this question, we need to know the basic theory related to the chapter limits. As we know, a given function is a $2x + 5$, which is a straight-line equation and it is continuous in nature. when we put $x \to 7$, we will get a determinate form. Thus, we need to simply put the value and we will get its approaching value or limit value.

Complete step by step answer: As we know, $2x + 5$ is an equation of straight line, which is continuous in nature.
And we know
A function is said to be continuous at a particular point if the follow
$\mathop {\lim }\limits_{x \to {a^ + }} {\text{f}}\left( x \right) = \mathop {\lim }\limits_{x \to {a^ - }} {\text{f}}\left( x \right) = \mathop {\lim }\limits_{x \to a} {\text{f}}\left( x \right)$
Here, ${\text{f}}\left( x \right) = 2x + 5$, and $a = 7$
we put these values and further proceed-
$\mathop {\lim }\limits_{x \to {7^ + }} \left( {2x + 5} \right) = \mathop {\lim }\limits_{x \to {7^ - }} \left( {2x + 5} \right) = \mathop {\lim }\limits_{x \to 7} \left( {2x + 5} \right)$
Hence, we say
$\mathop {\lim }\limits_{x \to {7^ - }} \left( {2x + 5} \right) = \mathop {\lim }\limits_{x \to 7} \left( {2x + 5} \right)$
= $\left( {2 \times 7 + 5} \right)$
= $\left( {14 + 5} \right)$
= $\left( {19} \right)$
Therefore, the limit of $2x + 5$as $x \to {7^ - }$ is 19.

Note:
A function may approach two different limits. One where the variable approaches its limit through values larger than the limit and the other where the variable approaches its limit through values smaller than the limit. In such a case, the limit is not defined but the right and left-hand limit exist.