Evaluate $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \sqrt[x^3]{1-x+\sin{x}}}$

In this limit problem, $x$ is a variable and also represents angle of the right triangle. They formed a special function in which algebraic and trigonometric functions are involved in exponential form. Now, let’s calculate the limit of the exponential function as $x$ approaches zero.

$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \sqrt[x^3]{1-x+\sin{x}}}$

Convert the function into Exponential Limit rule

The limit of the exponential form algebraic trigonometric function can be expressed in the form of limit rule of the exponential function $(1+x)^\frac{1}{x}$ as $x$ approaches zero. So, express the $x$ cubed root in exponent form.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Big(1-x+\sin{x}\Big)}^\dfrac{1}{x^3}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Big(1+\sin{x}-x\Big)}^\dfrac{1}{x^3}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Big(1+(\sin{x}-x)\Big)}^{\dfrac{1}{x^3} \Large{\times 1}}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Big(1+(\sin{x}-x)\Big)}^{\dfrac{1}{x^3} \Large{\times} \normalsize \dfrac{\sin{x}-x}{\sin{x}-x}}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Big(1+(\sin{x}-x)\Big)}^{\dfrac{1}{\sin{x}-x} \Large{\times} \normalsize \dfrac{\sin{x}-x}{x^3}}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Bigg[{\Big(1+(\sin{x}-x)\Big)}^{\dfrac{1}{\sin{x}-x}}\Bigg]}^\dfrac{\sin{x}-x}{x^3}}$

Now, use the limit rule of exponential function to find its value by calculating the limit of the base raised to the power of limit of exponent.

$= \,\,\,$ ${\Bigg[\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Big(1+(\sin{x}-x)\Big)}^{\dfrac{1}{\sin{x}-x}}}\Bigg]}^{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}-x}{x^3}}}$

If $x \to 0$, then $\sin{x} \to 0$. Therefore, $\sin{x}-x \to 0$. Therefore, if $x$ approaches zero, then $\sin{x}-x$ also approaches zero.

$= \,\,\,$ ${\Bigg[ \displaystyle \large \lim_{\sin{x}-x \,\to\, 0}{\normalsize {\Big(1+(\sin{x}-x)\Big)}^{\dfrac{1}{\sin{x}-x}} }\Bigg]}^{ \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}-x}{x^3}}}$

Evaluate limit of the base function

Take $y = \sin{x}-x$. Convert the function in base position in terms of $y$ but keep the limit of function in exponent position in terms of $x$.

$= \,\,\,$ ${\Bigg[ \displaystyle \large \lim_{y \,\to\, 0}{\normalsize {\Big(1+y\Big)}^{\dfrac{1}{y}}} \Bigg]}^{ \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}-x}{x^3}} }$

According to limit rule of exponential function, the limit of $1+y$ raised to the power of $1$ by $y$ as $y$ approaches zero is equal to $e$.

$= \,\,\,$ $e^{\, \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}-x}{x^3}}}$

Evaluate limit of the exponent function

$= \,\,\,$ $e^{\, \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}-x}{x^3}}}$

We have already evaluated the limit of $\dfrac{x-\sin{x}}{x^3}$ as x approaches zero previously.

$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{6}$

$\implies$ $-\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $-\dfrac{1}{6}$

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}-x}{x^3}}$ $\,=\,$ $-\dfrac{1}{6}$

Therefore, substitute the limit of the trigonometric function in the exponent position.

$= \,\,\,$ $e^{\, -\frac{1}{6}}$

$= \,\,\,$ $\dfrac{1}{e^{\, \frac{1}{6}}}$

$= \,\,\,$ $\dfrac{1}{\sqrt[\displaystyle 6]{e}}$