Question

Determine the constants a and b such that \[\lim_{x \rightarrow 0}\frac{ \sqrt[3]{ax+b}-2 }{ x }=\frac{ 5 }{ 4 }\]
(Without the use of L'Hospital's Rule)

Answer 1

hint:
using Taylor's expansion, around \(x=0\), I get this:
\[\Large f\left( x \right) = \frac{{\sqrt[3]{b} - 2}}{x} + \frac{a}{{3\sqrt[3]{{{b^2}}}}} + o\left( x \right)\]
where \(\large o(x)\) is a quantity which goes to zero as \(x\).
So can we hope to have the limit, if the two subsequent conditions hold:
\[\Large \left\{ \begin{gathered}
\sqrt[3]{b} - 2 = 0 \hfill \\
\hfill \\
\frac{a}{{3\sqrt[3]{{{b^2}}}}} = \frac{5}{4} \hfill \\
\end{gathered} \right.\]

Answer 2

oops.. we can hope...

Answer 3

\[
\begin{align*}
&\phantom{{}={}}\frac{\sqrt[3]{ax+b}-2}{x}\\
&u=ax+b\\
&x=\frac{u-b}{a}\\
&=\frac{\sqrt[3]{u}-2}{\frac{u-b}{a}}\\
&=a\left(\frac{\sqrt[3]{u}-2}{u-b}\right)
\end{align*}
\]
Can we evaluate this?
\[
\lim_{u\to b}\frac{\sqrt[3]{u}-2}{u-b}
\]

Answer 4

Is the limit finite? \(\displaystyle \lim_{x \rightarrow 0}\frac{ \sqrt[3]{ax+b}-2 }{ x }\) does not look finite to me since \(\sqrt[3]{ax+b}-2\) will not blow up to infinity but \(\frac{1}{x}\) will.

Answer 5

Taking x to infinite might make more sense.

Answer 6

\[
\begin{align*}
&\phantom{{}={}}\lim_{u\to {\color{red}\infty}}\frac{\sqrt[3]{u}-2}{u-b}\\
&=\lim_{u\to {\infty}}\frac{u^{-2/3}-2/u}{1-b/u}\quad\text{Divide up and down by }u\\
\end{align*}
\]
Can this be evaluated?

Answer 7

http://www.wolframalpha.com/input/?i=Limit [%28%28ax%2Bb%29^%281%2F3%29-2%29%2Fx%2C+x-%3E0]

It shows that the limit of x to 0 doesn't exist. If the link doesn't work, input this into Wolfram|Alpha.
Limit[((ax+b)^(1/3)-2)/x, x->0]

Answer 8

\(\lim\limits_{x \rightarrow 0}\dfrac{ \sqrt[3]{ax+b}-2 }{ x }\\~\\
=\lim\limits_{x \rightarrow 0}\dfrac{ \sqrt[3]{ax+b}-\sqrt[3]{8} }{ x }\\~\\

=\lim\limits_{x \rightarrow 0}\dfrac{ax+b-8 }{ x( \sqrt[3]{(ax+b)^2}+ \sqrt[3]{ax+b}\sqrt[3]{8} + \sqrt[3]{8^2} ) }\\~\\
\)

It's tricky, but its not so hard to convince ourselves that \(b=8\) and \(a=15\)... I'll try and come up with some reasoning why they must work..

Answer 9

How does this even work? This is indeed the correct answer as verified by Wolfram|Alpha. Why does it not blow up to infinity?

http://www.wolframalpha.com/input/?i=Limit [%28%2815x%2B8%29^%281%2F3%29-2%29%2Fx%2C+x-%3E0]

Limit[((15x+8)^(1/3)-2)/x, x->0]

Answer 10

I guess you could divide up and down by x?

Answer 11

http://www.wolframalpha.com/input/?i=Limit+%5B%28%28ax%2Bb%29%5E%281%2F3%29-2%29%2Fx%2C+x-%3E0%5D&a=%5E_Real

Answer 12

Ahh that looks like a good idea since we don't have the luxury to use L'hospitals...

\(\lim\limits_{x \rightarrow 0}\dfrac{ \sqrt[3]{ax+b}-2 }{ x }\\~\\
=\lim\limits_{x \rightarrow 0}\dfrac{ \sqrt[3]{ax+b}-\sqrt[3]{8} }{ x }\\~\\

=\lim\limits_{x \rightarrow 0}\dfrac{ax+b-8 }{ x( \sqrt[3]{(ax+b)^2}+ \sqrt[3]{ax+b}\sqrt[3]{8} + \sqrt[3]{8^2} ) }\\~\\

=\lim\limits_{x \rightarrow 0}\dfrac{a+(b-8)/x }{ \sqrt[3]{(ax+b)^2}+ \sqrt[3]{ax+b}\sqrt[3]{8} + \sqrt[3]{8^2} }\\~\\
= \text{some finite number} + \lim\limits_{x \rightarrow 0}\dfrac{b-8 }{ x(\text{some finite number}) }\\~\\
\)

then i hope we may argue that the overall limit exists only if \(b=8\)

Answer 13

If b neq 8 then 1/x blows up to infinity. If b=8 then b-8=0 and the second term vanishes?

Answer 14

Now how to determine a?

Answer 15

we may simply plugin b=8, evaluate the limit and set it equal to 5/4

Answer 16

In order for the limit to exist top will need to approach 0 since the bottom is approaching 0.
So \[\sqrt[3]{a(0)+b}-2=0 \\ \text{ gives us } b=8\]
plug back in and then I would apply the substitution to evaluate the limit (pretending I don't already know the limit)
\[u=\sqrt[3]{ax+b} \\ x \rightarrow 0 \text{ so }u \rightarrow 2 \\ \lim_{u \rightarrow 2} \frac{u-2}{\frac{u^3-8}{a}}\]
u^3-8 is easy to factor
from here it should be easy to come up with a

Answer 17

I'm kind of confused about which process would work out for this problem. Would the taylor expansion one work? Based on the work shown on this thread, it looks like we would get a=15 and b=8

Answer 18

@hpfan101, HP as in Harry Potter?

Answer 19

Yeah

Answer 20

I think there have been a couple of ways mentioned @hpfan101

Answer 21

Forgive if I don't mention your way... I see three ways.
Taylor
Rationalizing the numerator
A substitution

Answer 22

If you're not allowed to use L'Hopital's rule, you're generally not allowed to use Taylor series either @Michele_Laino

Answer 23

Oh, ok. I see the three ways now.

Answer 24

@Kainui Oh, so Taylor expansion isn't a method I could use for this problem?