The trigonometric function $1+\sin{x}$ and algebraic function $\dfrac{1}{x}$ formed a special function in exponential notation. The limit of this special function has to evaluate as $x$ approaches zero in this limit problem.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Big(1+\sin{x}\Big)^{\huge \frac{1}{x}}}$
The limit of special function in exponential form is similar to the limit rule of $(1+x)^{\large \frac{1}{x}}$ as $x$ approaches $0$. So, we have to adjust the given function same as this limit rule in this method for evaluating the limit of given function as $x$ approaches $0$.
A $\sin{x}$ function is required in exponent position. So, let’s try an acceptable adjustment.
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Big(1+\sin{x}\Big)^{\Large 1 \times \huge \frac{1}{x}}}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Big(1+\sin{x}\Big)^{\huge \frac{\sin{x}}{\sin{x}} \Large \times \huge \frac{1}{x}}}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Big(1+\sin{x}\Big)^{\huge \frac{1}{\sin{x}} \Large \times \huge \frac{\sin{x}}{x}}}$
It can be simplified as follows by using the power rule of exponents.
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\Big(1+\sin{x}\Big)^{\huge \frac{1}{\sin{x}}}\Bigg)^{\huge \frac{\sin{x}}{x}}}$
The limit of this special exponential function can be simplified by the exponentiation rule of the limits.
$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Big(1+\sin{x}\Big)^{\huge \frac{1}{\sin{x}}}\Bigg)^{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \frac{\sin{x}}{x}}}}$
The limit of the function in exponent position expresses a limit rule. According to the trigonometric limit rules, the limit of sinx/x as x approaches 0 is equal to one.
$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Big(1+\sin{x}\Big)^{\huge \frac{1}{\sin{x}}}\Bigg)^{\Large 1}}$
$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Big(1+\sin{x}\Big)^{\huge \frac{1}{\sin{x}}}\Bigg)}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Big(1+\sin{x}\Big)^{\huge \frac{1}{\sin{x}}}}$
Let $x \,\to\, 0$, then $\sin{x} \,\to\, \sin{0}$. Therefore, $\sin{x} \,\to\, 0$. It clears that $\sin{x}$ approaches zero as $x$ tends to $0$.
$=\,\,\,$ $\displaystyle \large \lim_{\sin{x} \,\to\, 0}{\normalsize \Big(1+\sin{x}\Big)^{\huge \frac{1}{\sin{x}}}}$
Assume, $y = \sin{x}$ and express the whole expression in terms of $y$.
$=\,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Big(1+y\Big)^{\huge \frac{1}{y}}}$
As per the exponential standard limit rule, the limit of the exponential function as $y$ closer to $0$ is equal to mathematical constant $e$.
$=\,\,\, e$
In this method, the complexity in the given function can be simplified by the logarithms.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Big(1+\sin{x}\Big)^{\huge \frac{1}{x}}}$
Use the fundamental rule of logarithms to release the function from its complexity.
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize e^{\displaystyle \normalsize \log_e{\Big(1+\sin{x}\Big)^{\huge \frac{1}{x}}}}}$
Now, use the exponential rule of limits to simplify the function further.
$=\,\,\,$ $e^{\,\displaystyle \large \lim_{x \,\to\, 0}{\displaystyle \normalsize \log_e{\Big(1+\sin{x}\Big)^{\huge \frac{1}{x}}}}}$
The mathematical function can be simplified by using the power rule of logarithms.
$=\,\,\,$ $e^{\,\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\frac{1}{x} \times \displaystyle \normalsize \log_e{\Big(1+\sin{x}\Big)\Bigg)}}}$
It can be simplified further by the logarithmic limit rule. A $\sin{x}$ function is required in the denominator of the logarithmic function. So, let us take a step to make the $\sin{x}$ to appear in the denominator of the log function.
$=\,\,\,$ $e^{\,\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(1 \times \frac{1}{x} \times \displaystyle \normalsize \log_e{\Big(1+\sin{x}\Big)\Bigg)}}}$
$=\,\,\,$ $e^{\,\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\frac{\sin{x}}{\sin{x}} \times \frac{1}{x} \times \displaystyle \normalsize \log_e{\Big(1+\sin{x}\Big)\Bigg)}}}$
$=\,\,\,$ $e^{\,\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\frac{1}{x} \times \frac{\sin{x}}{\sin{x}} \times \displaystyle \normalsize \log_e{\Big(1+\sin{x}\Big)\Bigg)}}}$
$=\,\,\,$ $e^{\,\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\frac{1 \times \sin{x}}{x} \times \displaystyle \normalsize \frac{\log_e{\Big(1+\sin{x}\Big)}}{\sin{x}}\Bigg)}}$
$=\,\,\,$ $e^{\,\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\frac{\sin{x}}{x} \times \displaystyle \normalsize \frac{\log_e{\Big(1+\sin{x}\Big)}}{\sin{x}}\Bigg)}}$
Now, use the product rule of limits to get the limit of product of functions by the product of their limits.
$=\,\,\,$ $e^{\Bigg(\,\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \frac{\sin{x}}{x}} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \frac{\log_e{\Big(1+\sin{x}\Big)}}{\sin{x}}\Bigg)}}$
As per the trigonometric limit rules, the limit of $\dfrac{\sin{x}}{x}$ as $x$ approaches $0$ is equal to one.
$=\,\,\,$ $e^{\Bigg(\, \displaystyle 1 \times \large \lim_{x \,\to\, 0}{\normalsize \frac{\log_e{\Big(1+\sin{x}\Big)}}{\sin{x}}\Bigg)}}$
$=\,\,\,$ $e^{\,\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \frac{\log_e{\Big(1+\sin{x}\Big)}}{\sin{x}}\Bigg)}}$
When $x \,\to\, 0$, the $\sin{x} \,\to\, \sin{0}$. Therefore, $\sin{x} \,\to\, 0$. Therefore, $\sin{x}$ approaches zero as $x$ closer to $0$.
$=\,\,\,$ $e^{\,\Bigg(\displaystyle \large \lim_{\sin{x} \,\to\, 0}{\normalsize \frac{\log_e{\Big(1+\sin{x}\Big)}}{\sin{x}}\Bigg)}}$
Now, take $z = \sin{x}$ and transform the whole function in terms of $z$.
$=\,\,\,$ $e^{\,\Bigg(\displaystyle \large \lim_{z \,\to\, 0}{\normalsize \frac{\log_e{\Big(1+z\Big)}}{z}\Bigg)}}$
As per the standard logarithmic rule, the limit of ln(1+x) by x as x approaches 0 is equal to one. Therefore, the limit of $\dfrac{\log_e{\Big(1+z\Big)}}{z}$ as $z$ approaches zero is also equal to one.
$=\,\,\,$ $e^{\,(1)}$
$=\,\,\,$ $e^1$
$=\,\,\,$ $e$
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