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I know how to solve this limit with the L'Hospital rule:

$$\lim_{x \rightarrow 4} \frac{(2x)^{1/3} - 2}{\sqrt{x} -2}$$

The answer is $\frac{2}{3}$. I am trying to solve it without the L'Hospital rule.

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To simplify things, substitute $x = 4y^6$. As $y \to 1$ we also have $x \to 4$.

Thus, $\displaystyle\lim_{x \to 4}\dfrac{(2x)^{1/3}-2}{\sqrt{x}-2} = \lim_{y \to 1}\dfrac{(2 \cdot 4y^6)^{1/3}-2}{\sqrt{4y^6}-2} = \lim_{y \to 1}\dfrac{2y^2-2}{2y^3-2}$.

Then factor the numerator and denominator and cancel common factors.

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hint: $a^3 -b^3 = (a-b)(a^2 +ab +b^2)$ and lots of terms cancel out.

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Hint: Make the change of variables $x=y^6$ and then factorize. We choose $6$ as the exponent of $y$ as we see that there we have to deal with $x^{(1/3)}$ and $\sqrt x$.

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Brute-force L'Hospital buster: Substitute $h=\sqrt{x}-2$. We then get $x=(2+h)^2$ and thus $$ \lim_{x \rightarrow 4} \frac{(2x)^{1/3} - 2}{\sqrt{x} -2} = \lim_{h\rightarrow 0} \frac{(2(2+h)^2)^{1/3} - 2}{h} = \lim_{h\rightarrow 0} \frac{\sqrt[3]{2}(2+h)^{2/3} - 2}{h} $$ which is, by definition, the derivative of $\sqrt[3]{2}\,t^{2/3}$ at $t=2$. Differentiate that symbolically and evaluate.

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$$\frac{(2x)^{1/3}-2}{\sqrt x-2}=\sqrt[3]2\frac{\sqrt[3]x-\sqrt[3]{2^2}}{\sqrt x-2}=\sqrt[3]2\frac{\overbrace{x-4}^{=(\sqrt x-2)(\sqrt x+2)}}{x^{2/3}+\sqrt[3]{4x}+4^{2/3}}\frac1{\sqrt x-2}=$$

$$=\frac{\sqrt[3]2(\sqrt x+2)}{x^{2/3}+\sqrt[3]{4x}+4^{2/3}}\xrightarrow[x\to 4]{}\frac{\sqrt[3]2\cdot4}{3\cdot2\sqrt[3]2}=\frac23$$

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You already get many answers, so I shall try something different hoping that you know Taylor series (if this is not the case, just discard my answer).

Built at $x=b$, you have by Taylor $$x^a=b^a+a b^{a-1} (x-b)+\frac{1}{2} (a-1) a b^{a-2} (x-b)^2+O\left((x-b)^3\right)$$ So, the expansion of the numerator is $$\frac{x-4}{6}-\frac{1}{72} (x-4)^2+O\left((x-4)^3\right)$$ and for the denominator $$\frac{x-4}{4}-\frac{1}{64} (x-4)^2+O\left((x-4)^3\right)$$ Performing the long division, you will the get that, close to $x=4$$$\frac{(2x)^{1/3} - 2}{\sqrt{x} -2}\approx \frac{2}{3}-\frac{x-4}{72}$$ which shows both the limit and how it is approached.

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  • $\begingroup$ Great answer! Thanks ! $\endgroup$ – math student Jan 10 '15 at 6:53
  • $\begingroup$ You are very welcome ! To me, Taylor series are extremely useful for limits. $\endgroup$ – Claude Leibovici Jan 10 '15 at 7:00

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