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Find a and b such $$\lim_{x\to 0}\frac{(ax+b)^\frac{1}{3} - 2}{x} = \frac{5}{12}$$

My Solution : I know that $b$ has to be $8$. In order to have $\displaystyle \frac{0}{0}$. But for finding a, I have to factorize and cross out $x$, which I didn't get how. I've done this by applying L'hopital rule and I get $a$ is 5. But I don't know how to do it without L'hopital rule. Can someone help me? Thanks.

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2 Answers 2

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Let $ax+8 = A^3$ and $2=B\;,$ Then $$\displaystyle ax = A^3-B^3\Rightarrow x=\frac{A^3-B^3}{a}$$

So we get $$\displaystyle \lim_{x\rightarrow 0}\frac{(A-B)\cdot a}{(A^3-B^3)} = \lim_{x\rightarrow 0}\frac{a}{A^2+B^2+AB} = \lim_{x\rightarrow 0}\frac{a}{(ax+8)^{\frac{2}{3}}+4+2\cdot (ax+8)^{\frac{1}{3}}}$$

So we get $$\displaystyle \frac{a}{12} = \frac{5}{12}$$ (Given)

So we get $a = 5$

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  • $\begingroup$ Wow this one explains to me a lot. Thanks $\endgroup$ Commented Oct 14, 2015 at 17:54
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Use the taylor series

$$(ax+8)^{\frac{1}{3}}=2+\frac{ax}{12}+O(x^2)$$

to get $a=5$.

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    $\begingroup$ $$(ax+b)^r=b^r+rab^{r-1}x+O(x^2)$$ This is analogue to $(a+b)^n=a^n+nab^{n-1}+...$ $\endgroup$
    – Peter
    Commented Oct 14, 2015 at 18:00
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    $\begingroup$ You could also calculate the derivate at $x=0$. The chain rule gives $((ax+8)^{\frac{1}{3}}) '=a\times \frac{1}{3} (ax+8)^{\frac{-2}{3}}$, so $f '(0)=\frac{1}{12}$ and we have $f(x)=f(0)+f'(0)\times x+O(x^2)=2+\frac{1}{12}a\times x+O(x^2)$ $\endgroup$
    – Peter
    Commented Oct 14, 2015 at 18:02
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    $\begingroup$ In general, the taylor series at the point $x=0$ is $f(x)=\sum_{k=0}^{\infty} \frac{f^k(x)x^k}{k!}$, if this series converges. $f^k(0)$ is the $k$-th derivate at $x=0$. $\endgroup$
    – Peter
    Commented Oct 14, 2015 at 18:07
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    $\begingroup$ I made an error in the post above : It must be $(a+b)^n=b^n+nab^{n-1}+...$ $\endgroup$
    – Peter
    Commented Oct 14, 2015 at 18:29
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    $\begingroup$ And the taylor series is $f(x)=\sum_{k=0}^\infty \frac{f^k(0)x^k}{k!}$ $\endgroup$
    – Peter
    Commented Oct 14, 2015 at 18:32

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