How can i solve a limit by precise definition when it has multiple factors on the numerator and a factor in the denominator?

Question

The limit that i'm trying to solve is this one

$$\lim_{{x\to1}}\frac{3x^2+2x+1}{2x-1}={6}$$

I don't know how to handle the case when two factors are in the numerator and one in the denominator

To find the value of $\delta > 0$ in the $\epsilon$-$\delta$ definition of a limit, you typically replace $|f(x) - L|$ in the inequality with $\epsilon$. In the provided expression:

$${|\frac{3x^2+2x+1}{2x-1}-6|}$$

$$\frac{|3x - 7||x - 1|}{|2x - 1|} < \epsilon$$

Answer 1

The limit that i'm trying to solve is this one $$\lim_{{x\to1}}\frac{3x^2+2x+1}{2x-1}={6}$$

The problem becomes trivial if you are allowed to use the following theorem:

Given:

• $~\displaystyle \lim_{x\to a} f(x) = M.$

• $~\displaystyle \lim_{x\to a} g(x) = N \neq 0.$

Then, $~\displaystyle \lim_{x\to a} \frac{f(x)}{g(x)} = \frac{M}{N}.$

Therefore, I am going to assume that the above approach is off-limits and it is required to go back to basics and construct an $~\epsilon, ~\delta~$ proof.

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$\underline{\text{Preliminary Ideas}}$

• Given any function $~f(x),~$ and any positive number $~a,~$ you will have that
  $|f(x)| < a \iff -a < f(x) < a.$

• Suppose $~0 < a < b,~$ and $~0 < c < d.~$
  Further suppose that $~a < f(x) < b,~$ and $~c < g(x) < d.$
  Then $~\displaystyle \frac{a}{d} < \frac{f(x)}{g(x)} < \frac{b}{c}.~$
  That is, a lower bound is achieved by combining a lower bound of the numerator and an upper bound for the denominator. Similarly, an upper bound is achieved by combining an upper bound of the numerator and a lower bound for the denominator.

• A common artificial method, in an $~\epsilon,\delta~$ proof, that ensures that all pertinent bounds are positive is to add a constraint to $~\delta,~$ such as (positive) $~\delta~$ must be $~\leq (1/10).$

• The previous bullet point is also handy when trying to relate (for example) $~\delta^2~$ to $~\delta.$ For example, if you are guaranteed that $~0 < \delta \leq (1/10),~$ then you know that $~\delta^2 \leq \dfrac{\delta}{10}.~$

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$\underline{\text{Formal Proof}}$

Let $~f(x) = 3x^2 + 2x + 1.~~$ Let $~g(x) = 2x - 1.~$

You are given a fixed $~\epsilon > 0.~$

You have to construct $~\delta,~$ based on $~\epsilon,~$

such that $~0 < |x-1| < \delta,~$ will guarantee that $~\displaystyle \left| ~\frac{f(x)}{g(x)} - 6 ~\right| < \epsilon.$

The $~\epsilon~$ constraint will be satisfied if and only if

$$6 - \epsilon < \frac{f(x)}{g(x)} < 6 + \epsilon. \tag1 $$

In addition to deriving some relationship between $~\delta~$ and $~\epsilon,~$ add the artificial constraint that $~\delta \leq (1/100).~$

So, you have that

$$0 < \delta \leq (1/100) ~~\text{and} ~~1-\delta < x < 1 + \delta. \tag2 $$

Now the problem reduces to deriving a relationship between $~\delta~$ and $~\epsilon~$ so that whenever the constraints in (2) above hold, the constraint in (1) above will also hold.

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Since $~1-\delta < x < 1 + \delta,~$
and since $~\delta^2 < \delta/100,~$
you have that

$(1 - \delta)^2 < x^2 < (1 + \delta)^2.$

Therefore, $~1 - 2\delta + \delta^2 < x^2 < 1 + 2\delta + \delta^2.~$

Therefore (rather loosely), $~1 - 3\delta < x^2 < 1 + 3\delta.~$

Therefore, $~3 - 9\delta < 3x^2 < 3 + 9\delta.$

Similarly, $~2 - 2\delta < 2x < 2 + 2\delta.$

Therefore

$$6 - 11\delta < f(x) < 6 + 11\delta.$$

Similarly,

$$1 - 2\delta < g(x) < 1 + 2\delta.$$

Therefore,

$$\frac{6 - 11\delta}{1 + 2\delta} < \frac{f(x)}{g(x)} < \frac{6 + 11\delta}{1 - 2\delta}.$$

Further,

$$\frac{6 - 11\delta}{1 + 2\delta} = \frac{6 + 12\delta}{1 + 2\delta} - \frac{23\delta}{1 + 2\delta} = 6 - \frac{23\delta}{1 + 2\delta} \\ > 6 - \frac{23\delta}{1} > 6 - 25\delta.$$

Similarly,

$$\frac{6 + 11\delta}{1 - 2\delta} = \frac{6 - 12\delta}{1 - 2\delta} + \frac{23\delta}{1 - 2\delta} = 6 + \frac{23\delta}{1 - 2\delta} \\ \leq 6 + \frac{23\delta}{0.98} < 6 + 25\delta.$$

Therefore,

$$6 - 25\delta < \frac{f(x)}{g(x)} < 6 + 25\delta.$$

So, let

$$\delta = \min\left( ~\frac{\epsilon}{25}, ~\frac{1}{100} ~\right).$$

This will imply that

$$6 - \epsilon \leq 6 - 25\delta ~~\text{and} ~~6 + 25\delta \leq 6 + \epsilon.$$

It will also imply that $~\delta \leq \dfrac{1}{100}.~$

So, by the analysis given, you will have that whenever $~0 < \delta \leq (1/100),~$ and $~|x - 1| < \delta,~$ you will have that

$$6 - 25\delta < \frac{f(x)}{g(x)} < 6 + 25\delta.$$

Answer 2

Prove that, $\lim_{{x\to1}}\frac{3x^2+2x+1}{2x-1}={6}$

Given, $\epsilon>0$

Choose, $\delta=\min\{\frac{1}3,\frac{\epsilon}{15}\}$

Suppose, $0<|x-1|<\delta$

Now, $0<|x-1|<\frac{1}3\Rightarrow\frac{2}3<x<\frac{4}3$

$\therefore 2-7<3x-7<4-7\Rightarrow -5<3x-7<-3\Rightarrow|3x-7|<5$

$\therefore \frac{4}3-1<2x-1<\frac{8}3-1\Rightarrow 3>\frac{1}{2x-1}>\frac{3}{5}\Rightarrow\frac{1}{|2x-1|}<3$

Check, $$|\frac{3x^2+2x+1}{2x-1}-{6}|{=\frac{|3x - 7||x - 1|}{|2x - 1|}\\<15\delta\\\le15\cdot\frac{\epsilon}{15}=\epsilon}$$

Answer 3

We want to show according to the definition of the limit \begin{align*} \lim_{x\to 1}\frac{3x^2+2x+1}{2x-1}=6 &\Longleftrightarrow\forall \varepsilon>0\ \exists\delta > 0:|x-1|<\delta,x\ne 1\\ &\qquad\qquad\Longrightarrow \left|\frac{3x^2+2x+1}{2x-1}-6\right|<\varepsilon \end{align*}

We have $|x-1|<\delta$ under control and can choose $\delta$ as small as we want. As OP has already shown it's a good idea to factor out $|x-1|$ from $|\frac{3x^2+2x+1}{2x-1}-6|$ since then we can use this factor to make the whole expression small (in fact smaller than $\varepsilon$).

We obtain \begin{align*} \color{blue}{\left|\frac{3x^2+2x+1}{2x-1}-6\right|}&\color{blue}{<\varepsilon}\\ \left|\frac{3x^2-10x+7}{2x-1}\right|&<\varepsilon\\ \left|\frac{3(x-1)^2-4x+4}{2x-1}\right|&<\varepsilon\\ \left|\frac{3(x-1)^2-4(x-1)}{2x-1}\right|&<\varepsilon\\ \color{blue}{\left|\frac{3x-7}{2x-1}\right||x-1|}&\color{blue}{<\varepsilon}\tag{1}\\ \end{align*}

In order to find a proper estimation for (1) we may WLOG assume that $x$ is in a small interval with center $1$, let's say \begin{align*} x\in\left[\frac{9}{10},\frac{11}{10}\right]\tag{2} \end{align*}

• Since $|3x-7|$ is decreasing in $\left[\frac{9}{10}, \frac{11}{10}\right]$ we get $|3x-7|\leq \left|3\cdot \frac{9}{10}-7\right|$.

• Since $\left|\frac{1}{2x-1}\right|$ is decreasing in $\left[\frac{9}{10}, \frac{11}{10}\right]$ we get $\left|\frac{1}{2x-1}\right|\leq \left|\frac{1}{2\cdot\frac{9}{10}-1}\right|$.

We can therefore use $\frac{9}{10}$ to estimate the expression in (1).

We obtain from (1) and (2) \begin{align*} \left|\frac{3x-7}{2x-1}\right||x-1|&\leq \left|\frac{3\cdot \frac{9}{10}-7}{2\cdot \frac{9}{10}-1}\right||x-1|=\frac{43}{8}|x-1|\tag{3} \end{align*} In order to get $\frac{43}{8}|x-1|<\varepsilon$ we set $\delta=\frac{4}{43}\varepsilon$ and finally get from (3) \begin{align*} \color{blue}{\left|\frac{3x-7}{2x-1}\right||x-1|}&\leq \frac{43}{8}|x-1| \leq \frac{43}{8}\cdot\frac{4}{43}\varepsilon =\frac{\varepsilon}{2} \color{blue}{<\varepsilon} \end{align*} and the claim follows.