Question

Calculate the following limits.
$\mathop {Lim}\limits_{x \to 0} \,\,\,\,\dfrac{{\sqrt[3]{{1 + \sin x}} - \sqrt[3]{{1 - \sin x}}}}{x}$.

Answer 1

Hint: To solve the question, we will determine whether the term after the limit is of the indiscriminate form or not. After deciding this, if the term becomes an indeterminate form after putting $x = 0$ then we will expand the terms in the numerator according to the expansion given below:
${\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right){x^2}}}{{2!}} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right){x^3}}}{{3!}} + .......$

Complete step-by-step solution -
We will put the limit $x = 0$ in the term given in question and determine whether the term is of indiscriminate form or not. So, now we will put $x = 0$. Thus we get the following:
$\mathop {Lim}\limits_{x \to 0} \,\,\,\,\dfrac{{\sqrt[3]{{1 + \sin 0}} - \sqrt[3]{{1 - \sin 0}}}}{0}$
$\mathop {Lim}\limits_{x \to 0} \,\,\,\,\dfrac{{\sqrt[3]{{1 + 0}} - \sqrt[3]{{1 - 0}}}}{0} = \dfrac{{1 - 1}}{0} = \dfrac{0}{0}$
So, after putting $x = 0$ in the above term, we get an indeterminate form of the form $\dfrac{0}{0}$. So, first thing we are going to do is to eliminate x from the denominator. For this, we will expand the terms given in numerator according to following relations:
${\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right){x^2}}}{{2!}} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right){x^3}}}{{3!}} + .......$
${\left( {1 - x} \right)^n} = 1 - nx + \dfrac{{n\left( {n - 1} \right){x^2}}}{{2!}} - \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right){x^3}}}{{3!}} + .......$
In our case, $x = \sin x$ and $n = 3$. Thus we get the following:
$\Rightarrow \mathop {Lim}\limits_{x \to 0} \,\,\,\,\dfrac{{{{\left( {1 + \sin x} \right)}^{\dfrac{1}{3}}} - {{\left( {1 - \sin x} \right)}^{\dfrac{1}{3}}}}}{x}$
$\Rightarrow \mathop {Lim}\limits_{x \to 0} \,\,\,\,\dfrac{{\left( {1 + \dfrac{{\sin x}}{3} + \dfrac{{\left( {\dfrac{1}{3}} \right)\left( {\dfrac{1}{3} - 1} \right){{\left( {\sin x} \right)}^2}}}{{2!}} + ......} \right) - \left( {1 - \dfrac{{\sin x}}{3} + \dfrac{{\left( {\dfrac{1}{3}} \right)\left( {\dfrac{1}{3} - 1} \right){{\left( {\sin x} \right)}^2}}}{{2!}} + ......} \right)}}{x}$
$\Rightarrow \mathop {Lim}\limits_{x \to 0} \,\,\,\,\dfrac{{\left( {1 + \dfrac{{\sin x}}{3} + \dfrac{{\dfrac{{ - 2}}{9}{{\sin }^2}x}}{2} + ......} \right) - \left( {1 - \dfrac{{\sin x}}{3} - \dfrac{{\dfrac{{ - 2}}{9}{{\sin }^2}x}}{{2!}} + ......} \right)}}{x}$
$\Rightarrow \mathop {Lim}\limits_{x \to 0} \,\,\,\dfrac{{\dfrac{{2\sin x}}{3} + \dfrac{{10{{\sin }^3}x}}{{81}} + ....}}{x}$
Now, here we can see that there will be infinite terms in the expansion of ${\left( {1 + x} \right)^n}$, but we have considered only the first few terms because the remaining terms will have higher power of $\sin x$ and when we will expand $\sin x$ it will still have higher power of x but we need only a single power of x because we need to eliminate only x from the denominator. So, here we can neglect high powers obtained above because that will anyway become equal to 0 after applying the limit. Thus, we have:
$\mathop {Lim}\limits_{x \to 0} \,\,\,\,\,\dfrac{2}{3}\left( {\dfrac{{\sin x}}{x}} \right)$
Now, we will expand $\sin x$ according to expansion given below:
$\sin x = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - .........$
Thus, we have:
$\Rightarrow \mathop {Lim}\limits_{x \to 0} \,\,\,\,\,\dfrac{{\dfrac{2}{3}\left( {x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - .........} \right)}}{x}$
$\Rightarrow \mathop {Lim}\limits_{x \to 0} \,\,\,\,\,\dfrac{2}{3}\left( {1 - \dfrac{{{x^2}}}{{3!}} + \dfrac{{{x^4}}}{{5!}} - .........} \right)$
Now, we will eliminate x from the denominator, so we will put $x = 0$ in all the terms. After doing this, we will get:
$\Rightarrow \mathop {Lim}\limits_{x \to 0} \,\,\,\,\,\dfrac{2}{3}\left( {1 - 0 + 0 - .........} \right)$
All the terms to the right of 1 will be zero. So we get:
$\Rightarrow \mathop {Lim}\limits_{x \to 0} \,\,\,\,\,\dfrac{2}{3}\left( 1 \right) = \dfrac{2}{3}$.

Note: The question can also be solved using L-Hospital’s rule because we have $\dfrac{0}{0}$ form. In this form, we differentiate numerator and denominator separately. Thus, the above question will also be solved as:
$\mathop {Lim}\limits_{x \to 0} \,\,\,\,\dfrac{{\sqrt[3]{{1 + \sin x}} - \sqrt[3]{{1 - \sin x}}}}{x}$
$\Rightarrow \mathop {Lim}\limits_{x \to 0} \,\,\,\,\dfrac{{\dfrac{d}{{dx}}\left( {\sqrt[3]{{1 + \sin x}} - \sqrt[3]{{1 - \sin x}}} \right)}}{{\dfrac{d}{{dx}}x}}$
$\Rightarrow \mathop {Lim}\limits_{x \to 0} \,\,\,\dfrac{{\dfrac{1}{3}{{\left( {1 + \sin x} \right)}^{\dfrac{{ - 2}}{3}}}\left( {\cos x} \right) - \dfrac{1}{3}{{\left( {1 - \sin x} \right)}^{\dfrac{{ - 2}}{3}}}\left( { - \cos x} \right)}}{1}$
$\Rightarrow \,\,\dfrac{{\dfrac{1}{3} + \dfrac{1}{3}}}{1}$
$\Rightarrow \dfrac{2}{3}$.