Question

How do you prove that the limit of \[\left( {\dfrac{{9 - 4{x^2}}}{{3 + 2x}}} \right) = 6\;\] as $x$ approaches -1.5 using the epsilon delta proof?

Answer 1

Hint:
Epsilon data proof:
In calculus, the $\varepsilon - \delta $ definition of a limit is an algebraically precise formulation of evaluating the limit of a function. The definition states that a limit $L$ of a function at a point ${x_0}$ exists if no matter how ${x_0}$ ​ is approached, the values returned by the function will always approach $L$.
Such that it can be written mathematically as:
Let $f\left( x \right)$ be a function defined on an open interval around ${x_0}$. Such that we can say that the limit of $f\left( x \right)$ as $x$ approaches ${x_0}$is\[L\].
$\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = L$
Now for every $\varepsilon > 0$ there exists $\delta > 0$ such that for all $x$.
$0 < \left| {x - {x_0}} \right| < \delta \Rightarrow \left| {f\left( x \right) - L} \right| < \varepsilon $
By using the above definitions and formulas we can solve the given question.

Complete step by step solution:
Given
The limit of \[\left( {\dfrac{{9 - 4{x^2}}}{{3 + 2x}}} \right) = 6\;\;{\text{\;as }}x{\text{ approaches }} - 1.5.......................\left( i \right)\]
So we need to prove the given statement using the epsilon delta proof.
For that we know that:
Now for every $\varepsilon > 0$ there exists $\delta > 0$ such that for all $x$.
$0 < \left| {x - {x_0}} \right| < \delta \Rightarrow \left| {f\left( x \right) - L} \right| < \varepsilon $
Here${x_0} = - 1.5$, so we can write:
\[0 < \left| {x + 1.5} \right| < \delta \;\mid \Rightarrow \left| {\left( {\dfrac{{9 - 4{x^2}}}{{3 + 2x}}} \right) - 6\;} \right| < \varepsilon .......................\left( {ii} \right)\]
Now in order to find the limit we need to manipulate the term \[\left| {\left( {\dfrac{{9 - 4{x^2}}}{{3 + 2x}}} \right) - 6\;} \right| < \varepsilon \] to find the value of\[\left| {x + 1.5} \right| < \]:
So we can write:
$
  \left| {\dfrac{{9 - 4{x^2}}}{{3 + 2x}} - 6} \right| = \left| {\dfrac{{\left( {3 - 2x} \right)\left( {3 + 2x} \right)}}{{3 + 2x}} - 6} \right| \\
   = \left| {\left( {3 - 2x} \right) - 6} \right| \\
   = \left| { - 2x - 3} \right|.........................\left( {iii} \right) \\
 $
Now we have\[\left| {x + 1.5} \right| < \delta \]:
Such that:
\[
  \left| {x + 1.5} \right| = \left| {x - \left( { - \dfrac{3}{2}} \right)} \right| \\
   = \left| { - 2} \right|\left| {x + \left( {\dfrac{3}{2}} \right)} \right| \\
   = 2\left| {x - \left( { - \dfrac{3}{2}} \right)} \right| \\
 \]
Now In order to make this less than $\varepsilon $, we have to make$\left| {x + \left( {1.5} \right)} \right|\;\;{\text{less than }}\dfrac{\varepsilon }{2}$ .
Now from (ii) we can write:
For every $\varepsilon > 0$ there exists $\delta > 0$ such that for all$x$.
\[0 < \left| {x + 1.5} \right| < \delta \;\mid \Rightarrow \left| {\left( {\dfrac{{9 - 4{x^2}}}{{3 + 2x}}} \right) - 6\;} \right| < \varepsilon \]
Also we have got
$0 < \left| {x + 1.5} \right| < \delta \; = \left| {x - \left( { - \dfrac{3}{2}} \right)} \right| < \delta $
Such that we can write:
\[
  \left| {\dfrac{{9 - 4{x^2}}}{{3 + 2x}} - 6} \right| = \left| {\dfrac{{\left( {3 - 2x} \right)\left( {3 + 2x} \right)}}{{3 + 2x}} - 6} \right| \\
   = \left| {\left( {3 - 2x} \right) - 6} \right| \\
   = \left| { - 2x - 3} \right| \\
   = \left| { - 2} \right|\left| {x - \left( { - \dfrac{3}{2}} \right)} \right| \\
   < 2\delta \\
   = 2\left( {\dfrac{\varepsilon }{2}} \right) \\
   = \varepsilon .....................................\left( {iv} \right) \\
 \]
We have shown that for any positive $\varepsilon $, there is a positive $\delta $ such that for all $x$ , if:
\[0 < \left| {x + 1.5} \right| < \delta \;\mid \Rightarrow \left| {\left( {\dfrac{{9 - 4{x^2}}}{{3 + 2x}}} \right) - 6\;} \right| < \varepsilon \]
Thereby using the basic definition of limit $\mathop {\lim }\limits_{x \to - 1.5} \left( {\dfrac{{9 - 4{x^2}}}{{3 + 2x}}} \right) = 6$
Hence proved the limit of \[\left( {\dfrac{{9 - 4{x^2}}}{{3 + 2x}}} \right) = 6\;\] as $x$ approaches -1.5 using the epsilon delta proof.

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.