Question

Need help with limits

Answer 1

Use the precise definition of the limit to prove\[\lim_{x \rightarrow 2}x^3+5x^2-10=18\]

Answer 2

You need to show that there exists a \(\delta \gt 0\) for any \(\epsilon \gt 0\) such that :
\[\large \left|x-2\right|\lt \delta \implies \left|(x^3+5x^2-10)-18\right| \lt \epsilon\]

Answer 3

yeah i got that, but i'm not sure what to do with the\[\left| x^3+5x^2-10-18 \right|\]

Answer 4

\[\lim_{x\to2}(x^3+5x^2-10)=18\]
You want to show that, given \(\epsilon>0\) and \(0<|x-2|<\delta\), we can have \(|x^3+5x^2-10-18|<\epsilon\).
\[\begin{align*}
|x^3+5x^2-10-18|&=|x^3+5x^2-28|
\end{align*}\]
I think the best way would be to first express this trinomial as a function of \(x-2\) instead of \(x\).
\[x^3+5x^2-28=a(x-2)^3+b(x-2)^2+c(x-2)+d\]
Expanding the RHS gives
\[ax^3-2ax^2+4ax-8a+bx^2-4bx+4b+cx-2c+d\]
and matching up your coefficients, you have
\[\begin{cases}
a=1\\
-2a+b=5\\
4a-4b+c=0\\
-8a+4b-2c+d=-28
\end{cases}\]

Answer 5

I factored it into (x-2)(x^2+7x+14)

Answer 6

Hmm did I make some silly mistake finding \(g(x-2)=f(x)-L\)?

Answer 7

\[x^3+5x^2-28=a(x-2)^3+b(x-2)^2+c(x-2)+d\]
Expanding the RHS gives
\[ax^3-\color{red}6ax^2+\color{red}{12}ax-8a+bx^2-4bx+4b+cx-2c+d\]
and matching up your coefficients, you have
\[\begin{cases}
a=1\\
-\color{red}6a+b=5\\
\color{red}{12}a-4b+c=0\\
-8a+4b-2c+d=-28
\end{cases}\]
I did!

This gives us
\[\begin{align*}|x^3+5x^2-28|&=|(x - 2)^3 + 11 (x - 2)^2 + 32 (x - 2)|\end{align*}\]

Answer 8

Right, so you factor the \(|x-2|\), giving you
\[|x^3+5x^2-28|=|x-2||(x-2)^2+11(x-2)+32|\]
so we have the desired \(|x-2|\) term that lets us use the hypothesized \(|x-2|<\delta\). To deal with the quadratic factor, we have to agree to set \(\delta\le1\), for example, then we know
\[\begin{align*}|x-2|<1~~\implies~~|(x-2)^2+11(x-2)+32|&\le|x-2|^2+11|x-2|+32\\
&<1+11+32\\
&=44
\end{align*}\]

Answer 9

i feel so lost lol

Answer 10

You want to prove the following "if-then" statement.
\[\textbf{If }\text{there exists }\delta\text{ such that }0<|x-2|<\delta,\\\quad\textbf{ then }\left|\left(x^3+5x^2-10\right)-18\right|<\epsilon\text{ for any }\epsilon>0\].
We want to be able to express the "then" inequality in terms of what we know, the "if" inequality. In doing so, we find an expression for \(\delta\) in terms of \(\epsilon\). \(\epsilon\) is a level of tolerance, you can think of it as a sort of error bound. If we can express \(\delta\) as a function of \(\epsilon\), we can establish the limit.

The general procedure is to manipulate the \(\epsilon\) inequality in such a way as to express it in terms of the \(\delta\) inequality. In doing so, we are essentially working backwards from our intended results (that so-and-so is less than \(\epsilon\)) to find a solution (a \(\delta\) that can be expressed in terms of \(\epsilon\)).

That's how I started off:
\[\left|\left(x^3+5x^2-10\right)-18\right|<\epsilon\]
Preliminary simplification and factoring:
\[\begin{align*}\left|\left(x^3+5x^2-10\right)-18\right|&=\left|x^3+5x^2-28\right|\\
&=\color{red}{|x-2|}\left|x^2+7x+14\right|
\end{align*}\]
Looking back at our "if", we have somthing to say about the red factor because we know already that \(|x-2|<\delta\):
\[\begin{align*}\color{red}{|x-2|}\left|x^2+7x+14\right|&<\color{red}\delta\left|x^2+7x+14\right|\end{align*}\]
We wanted to show that the LHS is less than \(\epsilon\), so we might want to say we're done and suppose
\[\delta\left|x^2+7x+14\right|=\epsilon~~\iff~~\delta=\frac{\epsilon}{\left|x^2+7x+14\right|}\]
but that won't work because \(\delta\) would be a function of both \(\epsilon\) and \(x\), and not just \(\epsilon\).

To get around this, we agree to set a constraint on \(\delta\); in my case, I assumed we have \(\delta\) be no more than 1, so that \(|x-2|<\delta\le1\). Given this assumption, we can determine a bound for the remaining quadratic term.

Since \(x^2+7x+14\) is irreducible (no real roots), we can't rely on factorizing. Instead, I suggested we write it in terms of \(x-2\). This will allow us to use the \(\delta\) bound we agree to.
\[\left|x^2+7x+14\right|=\left|(x-2)^2+11(x-2)+32\right|\]
The triangle inequality tells us
\[\left|x^2+7x+14\right|\le|x-2|^2+11|x-2|+32\]
Now we can use the \(\delta\) bound. Given that \(|x-2|<\delta\le1\), we then know that
\[\begin{align*}\left|x^2+7x+14\right|&\le|x-2|^2+11|x-2|+32\\&<\delta^2+11\delta+32\\&\le1^2+11\times1+32\\&=44\end{align*}\]
So, to recapitulate, we have
\[\begin{align*}\left|x^3+5x^2-28\right|&=|x-2|\left|x^2+7x+14\right|\\
&=|x-2|\left|(x-2)^2+11(x-2)+32\right|\\
&\le|x-2|\left(|x-2|^2+11|x-2|+32\right)\\
&<\delta\left(1^2+11+32\right)\\
&=\color{blue}{44\delta=\epsilon}~~\iff~~\delta=\frac{\epsilon}{44}
\end{align*}\]
This means the \(\delta\) we want to use will be
\[\delta=\min\left\{1,\frac{\epsilon}{44}\right\}\]