Question

How do you prove that the limit of \[3x + 5 = 35\] as x approaches 10 using the precise definition of a limit?

Answer 1

Hint:
We know that we say the limit of $f\left( x \right)$ is $L$ as $x$ approaches $a$, i.e.
$\mathop {\lim }\limits_{x \to a} f\left( x \right) = L$
Also we have to make $f\left( x \right)$ as close to $L$ and $x$ as close to $a$ from both the sides and not letting it be $a$.
So by using the above statement and expression we can solve the given question.

Complete step by step solution:
Given
$3x + 5............................\left( i \right)$
We have to prove here the limit of \[3x + 5 = 35\] as x approaches 10 using the precise definition of a limit.
Now on comparing the question and the statement) of limit we can say that $a = 10\,{\text{and}}\,\,{\text{f}}\left( x \right) = 3x + 5$. Since here ${\text{f}}\left( x \right) = 3x + 5$ it’s easy to find its limit directly since it’s an algebraic expression involving a single variable.
Such that to find the limit of ${\text{f}}\left( x \right) = 3x + 5$ we just have to find the value of ${\text{f}}\left( x \right) = 3x + 5$ at $x = a$.
Here we also know that $a = 10\,$, so we need to find the value of $f\left( x \right)$ at $x = 10$.
Such that we can write:
$
  f\left( a \right) = f\left( {10} \right) = 3\left( {10} \right) + 5 \\
   = 30 + 5 \\
   = 35.........................\left( {ii} \right) \\
 $
Therefore from (ii) we can write:
$
  L = \mathop {\lim }\limits_{x \to 10} \left( {3x + 5} \right) \\
   = 35.................................\left( {iii} \right) \\
 $
Hence proved that the limit of \[3x + 5 = 35\] as x approaches 10.

Note:
Also while approaching a question involving proofs one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.