limit problem involving a cube root and a square root

Question

I have been struggling with these problems for quite some time and cannot seem to find a solution, no matter what I try (u-substitution, rationalizing, etc.).

$$ \lim_{x \to\ 8} \frac{\sqrt[3]{8x}-\sqrt{x+8}}{x-8} \\ \lim_{x \to\ 1} \frac{\sqrt[3]{7+x^3}-\sqrt{3+x^2}}{x-1} $$ I am also struggling with this problem, it might be related: $\lim_{x\to\infty}\sqrt[3]{(x+1)^2}-\sqrt[3]{(x-1)^2} $

They all seem to be following a certain pattern, which I cannot understand.

Note: These problems should be solved by canceling out, i.e. without L'Hospital.

Answer 1

The identities

$$ a^3-b^3 = (a-b)(a^2+ab+b^2) $$ $$ a^2-b^2 = (a-b)(a+b) $$

will be used in this answer

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For the first limit, let $x= t^3$ $$ \lim = \lim_{t\to 2} \frac{2t - \sqrt{t^3+8}}{t^3 - 8} $$

Multiply by the conjugate of the numerator we get $$ \begin{align} \lim &= -\lim_{t\to 2} \frac{1}{2t+\sqrt{t^3+8}} \frac{t^3-4t^2+8}{t^3-8} \\ &= -\lim_{t\to 2} \frac{1}{2t+\sqrt{t^3+8}}\frac{(t-2)(t^2-2t-4)}{(t-2)(t^2+2t+4)} \\ &= -\lim_{t\to 2} \frac{1}{2t+\sqrt{t^3+8}}\frac{t^2-2t-4}{t^2+2t+4} \\ &= \frac{1}{24} \end{align} $$

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The second limit is a little more cumbersome. Let $u = 7 + x^3$ and $v = 3 + x^2$,so $$ \begin{align} \lim &= \lim_{x\to 1} \frac{u^{1/3} - v^{1/2}}{x-1} \\ &= \frac{1}{x-1}\frac{u - v^{3/2}}{u^{2/3}+u^{1/3}v^{1/2}+v} \\ &= \frac{1}{x-1}\frac{u^2 -v^3}{(u + v^{3/2})(u^{2/3}+u^{1/3}v^{1/2}+v)} \end{align} $$

Furthermore $$ \begin{align} u^2 - v^3 &= (7+x^3)^2 - (3+x^2)^3 \\ &= 22 - 27x^2 + 14x^3 - 9x^4 \\ &= -(x-1)(9x^3-5x^2+22x+22) \end{align} $$

Thus $$\lim = -\lim_{x\to 1 \\u\to 8\\v\to 4} \frac{9x^3-5x^2+22x+22}{(u + v^{3/2})(u^{2/3}+u^{1/3}v^{1/2}+v)} = -\frac{1}{4} $$

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For the last limit, let $x = 1/t$. Then

$$ \begin{align} \lim_{x\to\infty} &= \lim_{t\to 0} \frac{(1+t)^{2/3} - (1-t)^{2/3}}{t^{2/3}} \\ &= \lim_{t\to 0} \frac{(1+t)^2-(1-t)^2}{t^{2/3}((1+t)^{4/3}+(1+t)^{2/3}(1-t)^{2/3}+(1-t)^{4/3})} \\ &= \lim_{t\to 0} \frac{4t}{t^{2/3}((1+t)^{4/3}+(1+t)^{2/3}(1-t)^{2/3}+(1-t)^{4/3})} \\ &= \lim_{t\to 0} \frac{4t^{1/3}}{(1+t)^{4/3}+(1+t)^{2/3}(1-t)^{2/3}+(1-t)^{4/3}} \\ &= 0 \end{align} $$

Answer 2

You can write $u=x-8$ in the first and take the limit as $u \to 0$. You can then expand the roots in Taylor series. $$\begin {align} \lim_{x \to\ 8} \frac{\sqrt[3]{8x}-\sqrt{x+8}}{x-8}&=\lim_{u \to 0}\frac{\sqrt[3]{8(u+8)}-\sqrt{u+16}}{u}\\&=\lim_{u \to 0}\frac {\sqrt[3]{64}\sqrt[3]{1+\frac u8}-4\sqrt{1+\frac u{16}}}u\\&=\lim_{u \to 0}\frac {4(1+\frac u{24}+\ldots)-4(1+\frac u{32}+\ldots)}u\\&=\lim_{u \to 0}\frac 1{24}+\ldots \end{align}$$ where the dots show terms with more powers of $u$. We are expanding the roots in a Taylor series.

Answer 3

Both of these are, by definition, derivatives. $$ \begin{align} \lim_{x\to8}\frac{\left(\sqrt[3]{8x}-\sqrt{x+8}\right)-0}{x-8} &\equiv\left.\frac{\mathrm{d}}{\mathrm{d}x}\left(\sqrt[3]{8x}-\sqrt{x+8}\right)\right|_{x=8}\\ &=\left.\frac23x^{-2/3}-\frac12(x+8)^{-1/2}\right|_{x=8}\\ &=\frac1{24} \end{align} $$ and $$ \begin{align} \lim_{x\to1}\frac{\left(\sqrt[3]{7+x^3}-\sqrt{3+x^2}\right)-0}{x-1} &\equiv\left.\frac{\mathrm{d}}{\mathrm{d}x}\left(\sqrt[3]{7+x^3}-\sqrt{3+x^2}\right)\right|_{x=1}\\ &=\left.\vphantom{\frac{\mathrm{d}}{\mathrm{d}x}}x^2\left(7+x^3\right)^{-2/3}-x\left(3+x^2\right)^{-1/2}\right|_{x=1}\\ &=-\frac14 \end{align} $$

Answer 4

Not a terribly good idea, but feasible: write $$\sqrt[3]{8x} - \sqrt{x+8} = \sqrt[6]{(8x)^2} - \sqrt[6]{(x+8)^3}$$ and continue like in the Dylan's answer using the identity $$a^6 - b^6 = (a - b)(a^5 + a^4b + \cdots + ab^4 + b^5).$$