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Compute $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x }}{x}$

Original question was to solve $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+ x } - \sqrt[3]{1-x }}{x}$ and it was solved by adding and subtracting 1 in denominator. Making it in the form $\lim\limits_{x \to a}\dfrac {x^n - a^n}{x-a} = ax^{n-1}$

How to solve for above limit without using lhopitals rule?

4 Answers 4

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Hint. Use $$u-v=\frac{u^3-v^3}{u^2+uv+v^2}$$ with $$u=(1+\sin x)^{1/3}\ ,\quad v=(1-\sin x)^{1/3}\ .$$ You should know $$\lim_{x\to0}\frac{\sin x}x$$ without appealing to l'Hopital's Rule.

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    $\frac23$ realy. – 2017-02-14
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    I got ! Thanks :) I unconsciously cancelled one $1$ in denominator :P – 2017-02-14
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Just another way using Taylor series.

$$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ Now, using the generalized binomial theorem $$\sqrt[3]{1+\sin(x)}=1+\frac{x}{3}-\frac{x^2}{9}+\frac{x^3}{162}+O\left(x^4\right)$$ $$\sqrt[3]{1-\sin(x)}=1-\frac{x}{3}-\frac{x^2}{9}-\frac{x^3}{162}+O\left(x^4\right)$$ Using the above, $$\sqrt[3]{1+\sin(x)}-\sqrt[3]{1-\sin(x)}=\frac{2 x}{3}+\frac{x^3}{81}+O\left(x^4\right)$$ $$\dfrac{ \sqrt[3]{1+\sin (x)} - \sqrt[3]{1- \sin (x) }}{x}=\frac{2}{3}+\frac{x^2}{81}+O\left(x^3\right)$$ which shows the limit and also how it is approached.

Making the problem more general considering $$A=\frac{(1+\sin (x))^k-(1-\sin (x))^k}{x}$$ and using the same process, we should obtain $$A=2k+\frac{k \left(k^2-3 k+1\right)}{3} x^2+O\left(x^3\right)$$

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    Expansion at order $1$ would be enough. – 2017-02-14
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    I totally agree with you for the limit. I am always giving the next term on purpose. Have a look at http://matheducators.stackexchange.com/questions/8339/teaching-limits-and-asymptotics-at-the-same-time to see my concerns. Cheers. – 2017-02-14
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If you already know how to find the limit $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+ x } - \sqrt[3]{1-x }}{x}$ and also that $\dfrac{\sin x}x$ tends to 1, then you can simply use $$ \frac {\sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x}}{x} = \frac {\sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x}}{\sin x} \cdot \frac{\sin x}x. $$

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    Ultimately when $x \to 0$ then $\sin x ∼x$ and we can replace $\sin x$ with $x$ ? – 2017-02-14
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    That's true, but *only in some situations.* (Sometimes the terms of order $x^2, x^3, \dots$ are not negligible. Sometimes such replacement can change result completely.) – 2017-02-14
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$$\lim _{ x\to 0 }{ \frac { { (1+\sin x) }^{ \frac { 1 }{ 3 } }-{ (1-\sin x) }^{ \frac { 1 }{ 3 } } }{ x } } \frac { \left( { (1+\sin x) }^{ \frac { 2 }{ 3 } }+{ (1+\sin x) }^{ \frac { 1 }{ 3 } }{ (1-\sin x) }^{ \frac { 1 }{ 3 } }+{ (1-\sin x) }^{ \frac { 2 }{ 3 } } \right) }{ \left( { (1+\sin x) }^{ \frac { 2 }{ 3 } }+{ (1+\sin x) }^{ \frac { 1 }{ 3 } }{ (1-\sin x) }^{ \frac { 1 }{ 3 } }+{ (1-\sin x) }^{ \frac { 2 }{ 3 } } \right) } =\\ =\lim _{ x\to 0 }{ \frac { 1+\sin { x-1+\sin { x } } }{ x } \frac { 1 }{ \left( { (1+\sin x) }^{ \frac { 2 }{ 3 } }+{ (1+\sin x) }^{ \frac { 1 }{ 3 } }{ (1-\sin x) }^{ \frac { 1 }{ 3 } }+{ (1-\sin x) }^{ \frac { 2 }{ 3 } } \right) } } =\\ =\lim _{ x\to 0 }{ \frac { 2\sin { x } }{ x } \frac { 1 }{ \left( { (1+\sin x) }^{ \frac { 2 }{ 3 } }+{ (1+\sin x) }^{ \frac { 1 }{ 3 } }{ (1-\sin x) }^{ \frac { 1 }{ 3 } }+{ (1-\sin x) }^{ \frac { 2 }{ 3 } } \right) } } =\frac { 2 }{ 3 } \\ \\ $$

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    denominator + not - – 2017-02-14
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    Thanks , I accepted David answer because it was 2 minutes early, btw +1 – 2017-02-14
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    yes,of course you are right – 2017-02-14