Limit problem bee

Let $\sqrt[3]{1+\tan x}=A$ and $\sqrt[3]{1+\sin x}=B$ . $L=\lim_{x\to0}\frac{\sqrt[3]{1+\tan x}-\sqrt[3]{1+\sin x}}{x^3}=\lim_{x\to0}\frac{A-B}{x^3}$ Now, since $A^3-B^3=(A-B)(A^2+AB+B^2)$ : $\begin{aligned}L&=\lim_{x\to0}\frac{A-B}{x^3}\times\frac{A^2+AB+B^2}{A^2+AB+B^2}\\ &=\lim_{x\to0}\frac{A^3-B^3}{x^3}\times\frac1{A^2+AB+B^2}\\ &=\lim_{x\to0}\frac{(1+\tan x)-(1+\sin x)}{x^3}\times\frac1{A^2+AB+B^2}\end{aligned}$ When $x=0,\,A=1$ and $B=1$ . This means the right side is defined for $x=0$ , and we can remove it from the limit: $\begin{aligned}L&=\left(\lim_{x\to0}\frac{1+\tan x-1-\sin x}{x^3}\right)\times\frac1{1^2+1\times1+1^2}\\ L&=\frac13\lim_{x\to0}\frac{\tan x-\sin x}{x^3}\end{aligned}$ Now we simply use L'Hospital's rule 3 times: $\begin{aligned}L&=\frac13\lim_{x\to0}\frac{\sec^2 x-\cos x}{3x^2}\\ &=\frac13\lim_{x\to0}\frac{2\sec^2 x\tan x+\sin x}{6x}\\ &=\frac13\lim_{x\to0}\frac{2\sec^4 x+4\sec^2 x\tan^2 x+\cos x}6\\ &=\frac13\times\frac{2+0+1}6\\ &=\frac13\times\frac36\\ &=\frac16\end{aligned}$ Therefore $\boxed{k=6}$

Using Taylor expansions as $x\rightarrow 0$ :

$\displaystyle{(1+x)^n=1+nx+O(x)}$

$\displaystyle{\tan x=x+\frac{1}{3} x^3+O(x^3)}$

$\displaystyle{\sin x=x-\frac{1}{6} x^3+O(x^3)}$

The limit becomes

$\displaystyle{\lim_{x\rightarrow 0} \frac{1+\frac{1}{3}\tan x-\left( 1+\frac{1}{3}\sin x \right)}{x^3}}$

$=\displaystyle{\frac{1}{3}\lim_{x\rightarrow 0} \frac{\tan x-\sin x}{x^3}}$

$=\displaystyle{\frac{1}{3}\lim_{x\rightarrow 0} \frac{x+\frac{1}{3}x^3 -\left( x-\frac{1}{6}x^3 \right)}{x^3}}$

$=\displaystyle{\frac{1}{6}}$

We can use binomial expansion to solve the question

$\lim_{x\to 0}\frac{1+\frac{1}{3} \tan(x) -1-\frac{1}{3} \sin(x)}{x^3}$

and apply L' Hospital's rule 3 times

Observe that from the algebraic identity: $\begin{aligned}a^3 - b^3 &= (a-b)(a^2 + ab + b^2) \\ a-b &= \frac{a^3 - b^3}{a^2 + ab + b^2}\end{aligned}$ We could substitute $a = \sqrt[3]{1 + \tan{x}}$ and $b = \sqrt[3]{1 + \sin{x}}$ and by putting it into the limit, we get $\begin{aligned}\lim_{x\to 0} \frac{\sqrt[3]{1 + \tan{x}} - \sqrt[3]{1 + \sin{x}}}{x^3} &= \lim_{x\to 0} \frac{(1 + \tan{x}) - (1 + \sin{x})}{x^3}\cdot \frac{1}{\sqrt[\frac{3}{2}]{1 + \tan{x}} + \sqrt[3]{(1 + \sin{x})(1 + \tan{x})} + \sqrt[\frac{3}{2}]{1 + \sin{x}}} \\ &= \lim_{x\to 0} \frac{\tan{x} - \sin{x}}{x^3}\cdot \frac{1}{\sqrt[\frac{3}{2}]{1 + \tan{x}} + \sqrt[3]{(1 + \sin{x})(1 + \tan{x})} + \sqrt[\frac{3}{2}]{1 + \sin{x}}}\end{aligned}$ The Maclaurin series for $\sin{x}$ and $\tan{x}$ is $\begin{aligned} \sin{x} &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots \\ \tan{x} &= x + \frac{x^3}{3} + \frac{2x^5}{15} + \frac{17x^7}{315} + \dots \end{aligned}$ and using them to the left side fraction in the limit gives us $\begin{aligned} &= \lim_{x\to 0} \frac{\left(x + \frac{x^3}{3} + \frac{2x^5}{15} + \frac{17x^7}{315} + \dots\right) - \left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots \right)}{x^3}\cdot \frac{1}{\sqrt[\frac{3}{2}]{1 + \tan{x}} + \sqrt[3]{(1 + \sin{x})(1 + \tan{x})} + \sqrt[\frac{3}{2}]{1 + \sin{x}}} \\ &= \lim_{x\to 0} \frac{\frac{1}{2}x^3 + \left(\frac{2}{15} - \frac{1}{5!}\right)x^5 + \dots}{x^3}\cdot \frac{1}{\sqrt[\frac{3}{2}]{1 + \tan{x}} + \sqrt[3]{(1 + \sin{x})(1 + \tan{x})} + \sqrt[\frac{3}{2}]{1 + \sin{x}}} \\ &= \lim_{x\to 0} \left(\frac{1}{2} + \left(\frac{2}{15} - \frac{1}{5!}\right)x^2 + \dots \right)\cdot \frac{1}{\sqrt[\frac{3}{2}]{1 + \tan{x}} + \sqrt[3]{(1 + \sin{x})(1 + \tan{x})} + \sqrt[\frac{3}{2}]{1 + \sin{x}}} \end{aligned}$ We could substitute $x=0$ to the expression of the limit, it evaluates to $\frac{1}{6}$ and hence $\boxed{k=6}$ .

tan x = 1 + $\frac{x^3}{3}$ + $\frac{2x^5}{15}$ + ...

sin x = x - $\frac{x^3}{6}$ + $\frac{x^5}{120}$ - ...

(1 + tan x)^(1/3) - (1 + sin x)^(1/3) = $\frac{x^3}{6}$ + higher powers of x (using binomial expansion as |tan x| < 1 and |sin x| < 1 as x tends to 0)

Hence, the required limit is 1/6