The limit of sine of angle $x$ divided by $x$ whole raised to the power of the reciprocal of $x$ square should be evaluated in this problem as the value of $x$ approaches zero. The given function is in exponential notation. So, let’s try to find the limit by the power rule of limits.

$\implies$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{\sin{x}}{x}\Big)^{\dfrac{1}{x^2}}}$ $\,=\,$ $\Big(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{x}}{x}\Big)^{\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1}{x^2}}}}$

According to the trigonometric limit rule in sine function, the limit of sine of angle $x$ divided $x$ is one as the $x$ approaches $0$. Similarly, the limit of multiplicative inverse of $x$ square can be calculated by the direct substitution.

$\,\,=\,\,$ $(1)^{\dfrac{1}{0^2}}$

$\,\,=\,\,$ $1^{\dfrac{1}{0}}$

$\,\,=\,\,$ $1^\infty$

It is evaluated that the limit of sine of angle $x$ divided by $x$ whole raised to the power of one divided by $x$ square is one raised to the power of infinity and it is an indeterminate form.

$\implies$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{\sin{x}}{x}\Big)^{\dfrac{1}{x^2}}}$ $\,=\,$ $1^\infty$

The indeterminate form as the limit expresses that we should think about an alternative method to find the limit of this function in this problem. It is a hard limit question in calculus and let’s learn how to find the limit in fundamental approach.

The exponential form in the given function is the mainly reason for the indeterminate form the one raised to the power of infinity. So, it is a smart idea to convert the exponential function into other possible form.

Firstly, let’s denote the limit of the given function by a variable $y$.

$\implies$ $y$ $\,=\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{\sin{x}}{x}\Big)^{\dfrac{1}{x^2}}}$

Now, take the natural logarithm on both sides of the equation for releasing the function from exponential notation.

$\implies$ $\log_{e}{y}$ $\,=\,$ $\log_{e}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{\sin{x}}{x}\Big)^{\dfrac{1}{x^2}}\Bigg)}}$

Now, let us focus on releasing the function from exponential form on the right hand side of the equation.

$\,\,=\,\,$ $\log_{e}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{\sin{x}}{x}\Big)^{\dfrac{1}{x^2}}\Bigg)}}$

Use the composite limit rule to take out the limit operation from the logarithm and this step is useful to release the function from exponential form.

$\,\,=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \log_{e}{{\Big(\dfrac{\sin{x}}{x}\Big)^{\dfrac{1}{x^2}}}}}$

The function can be released from exponential form as per the fundamental power logarithmic identity.

$\,\,\,\therefore\,\,\,\,\,\,$ $\log_{e}{y}$ $\,=\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{1}{x^2} \times \log_{e}{\Big(\dfrac{\sin{x}}{x}\Big)\bigg)}}$

The given function in exponential form is converted as a product of two separate functions. One function is an algebraic function and the second function is a function in logarithmic operation. The limit of reciprocal function can be calculated by the direct substitution but the limit of logarithmic function should be calculated by the logarithmic limit rule.

So, the function inside the logarithm should be in the form of the logarithmic limit rule. Now, add one and subtract one inside the logarithm.

$\,\,=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{1}{x^2} \times \log_{e}{\Big(1-1+\dfrac{\sin{x}}{x}\Big)\bigg)}}$

$\,\,=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{1}{x^2} \times \log_{e}{\Big(1+\dfrac{\sin{x}}{x}-1\Big)\bigg)}}$

$\,\,=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Bigg(\dfrac{1}{x^2} \times \log_{e}{\bigg(1+\Big(\dfrac{\sin{x}}{x}-1\Big)\bigg)\Bigg)}}$

$\,\,=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Bigg(\dfrac{1}{x^2} \times \log_{e}{\bigg(1+\Big(\dfrac{\sin{x}-x}{x}\Big)\bigg)\Bigg)}}$

The function inside the logarithm should also be there at outside of the logarithmic function for using the logarithmic limit rule. Now, it is time for an adjustment for making the function inside the logarithm to appear at outside of the function.

$\,\,=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Bigg(\dfrac{1}{x^2} \times 1 \times \log_{e}{\bigg(1+\Big(\dfrac{\sin{x}-x}{x}\Big)\bigg)\Bigg)}}$

$\,\,=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Bigg(\dfrac{1}{x^2} \times \dfrac{\Big(\dfrac{\sin{x}-x}{x}\Big)}{\Big(\dfrac{\sin{x}-x}{x}\Big)} \times \log_{e}{\bigg(1+\Big(\dfrac{\sin{x}-x}{x}\Big)\bigg)\Bigg)}}$

Now, split the second factor inside the limit operation as a product of two functions as per the multiplication of the fractions.

$\,\,=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Bigg(\dfrac{1}{x^2}}$ $\times$ $\Big(\dfrac{\sin{x}-x}{x}\Big)$ $\times$ $\dfrac{1}{\Big(\dfrac{\sin{x}-x}{x}\Big)}$ $\times$ $\log_{e}{\bigg(1+\Big(\dfrac{\sin{x}-x}{x}\Big)\bigg)\Bigg)}$

$\,\,=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Bigg(\dfrac{1}{x^2}}$ $\times$ $\dfrac{\sin{x}-x}{x}$ $\times$ $\dfrac{1}{\Big(\dfrac{\sin{x}-x}{x}\Big)}$ $\times$ $\log_{e}{\bigg(1+\Big(\dfrac{\sin{x}-x}{x}\Big)\bigg)\Bigg)}$

Multiply the first two factors to find their product. Similarly, multiply the remaining factors to find their product.

$\,\,=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Bigg(\dfrac{1 \times (\sin{x}-x)}{x^2 \times x} \times \dfrac{1 \times \log_{e}{\bigg(1+\Big(\dfrac{\sin{x}-x}{x}\Big)\bigg)}}{\Big(\dfrac{\sin{x}-x}{x}\Big)}\Bigg)}$

According to the product rule of exponents, the product of $x$ square and $x$ is $x$ cube.

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

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

The limit of the product of two functions can be calculated by the product of their limits as per the product rule of limits.

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

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

The math equation expresses that the natural logarithm of variable $y$ should be evaluated by finding the product after evaluating the limit of every function. Now, let us find the limit of every function to find the natural logarithm of variable $y$.

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

Take out the negative one from the numerator of rational function firstly to start the process of finding the limit of sine of angle $x$ minus $x$ divided by cube of variable $x$ as the value of $x$ is closer to zero.

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

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

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

We have already evaluated that the limit of $x$ minus sine of angle $x$ divided by $x$ cube is one divided by six as the value of $x$ is closer to zero.

$=\,\,$ $(-1) \times \dfrac{1}{6}$

$=\,\,$ $-\dfrac{1}{6}$

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

$(2).\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\log_{e}{\bigg(1+\Big(\dfrac{\sin{x}-x}{x}\Big)\bigg)}}{\Big(\dfrac{\sin{x}-x}{x}\Big)}}$

The logarithmic operator is involved in the rational function. So, the logarithmic limit rule can be used to find the limit of the function but it is required to do some acceptable settings.

The rational function inside the logarithmic function is also there in the denominator but the input of the limit operation should also be same. Let’s try to set the input of the limit operation.

$(1).\,\,\,$ If $x \,\to\, 0$, then $\sin{x} \,\to\, \sin{0}$. Therefore, $\sin{x} \,\to\, 0$

$(2).\,\,\,$ If $\sin{x} \,\to\, 0$, then $\sin{x}-x \,\to\, 0-x$. Now, $\sin{x}-x \,\to\, -x$ but the value of $x$ is closer to zero in this case. Therefore, $\sin{x}-x \,\to\, 0$

$(3).\,\,\,$ If $\sin{x}-x \,\to\, 0$, then $\dfrac{\sin{x}-x}{x} \,\to\, \dfrac{0}{x}$. Therefore, $\dfrac{\sin{x}-x}{x} \,\to\, 0$

According to the above three steps, when the value of $x$ approaches zero, the value of the ratio of sine of angle $x$ minus $x$ to the $x$ also tends to zero.

$\implies$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\log_{e}{\bigg(1+\Big(\dfrac{\sin{x}-x}{x}\Big)\bigg)}}{\Big(\dfrac{\sin{x}-x}{x}\Big)}}$ $\,=\,$ $\displaystyle \large \lim_{{\normalsize \dfrac{\sin{x}-x}{x}}\,\to\,0}{\normalsize \dfrac{\log_{e}{\bigg(1+\Big(\dfrac{\sin{x}-x}{x}\Big)\bigg)}}{\Big(\dfrac{\sin{x}-x}{x}\Big)}}$

Now, denote the sine of angle $x$ minus $x$ divided by $x$ by a variable $z$ for our convenience. It means, $z$ $\,=\,$ $\dfrac{\sin{x}-x}{x}$

$\,\,=\,\,$ $\displaystyle \large \lim_{z\,\to\,0}{\normalsize \dfrac{\log_{e}{\big(1+(z)\big)}}{(z)}}$

$\,\,=\,\,$ $\displaystyle \large \lim_{z\,\to\,0}{\normalsize \dfrac{\log_{e}{(1+z)}}{z}}$

$\,\,=\,\, 1$

According to the logarithmic limit rule, the limit of the natural logarithm of one plus $z$ divided by $z$ is one as the value of $z$ tends to zero.

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\log_{e}{\bigg(1+\Big(\dfrac{\sin{x}-x}{x}\Big)\bigg)}}{\Big(\dfrac{\sin{x}-x}{x}\Big)}}$ $\,=\,$ $1$

We have evaluated that

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

$(2).\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\log_{e}{\bigg(1+\Big(\dfrac{\sin{x}-x}{x}\Big)\bigg)}}{\Big(\dfrac{\sin{x}-x}{x}\Big)}}$ $\,=\,$ $1$

Now, substitute the limits in the equation to calculate the natural logarithm of $y$.

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

$\implies$ $\log_{e}{y}$ $\,=\,$ $\Big(-\dfrac{1}{6}\Big)$ $\times$ $1$

$\,\,\,\therefore\,\,\,\,\,\,$ $\log_{e}{y}$ $\,=\,$ $-\dfrac{1}{6}$

It is evaluated that the natural logarithm of $y$ is equal to negative one divided by six. Now, it is time to eliminate the logarithm from equation to find the value of variable $y$ and it can be done by the mathematical relationship between the exponents and logarithms.

$\implies$ $y$ $\,=\,$ $e^{\displaystyle -\small \dfrac{1}{6}}$

We have assumed that $y$ $\,=\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{\sin{x}}{x}\Big)^{\dfrac{1}{x^2}}}$

Now, replace the value of variable $y$ by its actual value in $x$ to find the limit of the given tough function.

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{\sin{x}}{x}\Big)^{\dfrac{1}{x^2}}}$ $\,=\,$ $e^{\displaystyle -\small \dfrac{1}{6}}$

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