How to solve the limit for power tending to infinity How does one find
$$
\lim_{x \to 0} \left[1 + x\sin(\pi - x) \right]^{1/x}\ ?
$$
Does the limit exist? If yes, how is it found?
 A: hint
$$f (x)=e^{\frac {1}{x}\ln (1+x\sin (x))}$$
but $$\ln (1+x\sin (x))\sim x^2 \;\;(x\to 0)$$
thus your limit is $e^0=1$.
or
$$f (x)=e^{ \sin (x)\frac {\ln (1+x\sin (x))}{x\sin (x)}}$$
and $$\lim_{X\to 0}\frac { \ln (1+X) }{  X}=1$$
A: $$
\begin{aligned}
\lim _{x\to 0}\left(1+x\sin \left(\pi-x\right)\right)^{\frac{1}{x}}
& = \lim _{x\to 0}\space\exp\left(\frac{\ln\left(1+x\sin \left(\pi-x\right)\right)}{x}\right)
\\& = \lim _{x\to 0}\space \exp\left(\frac{x^2-\frac{x^4}{6}+o\left(x^4\right)}{x}\right)
\\& = \exp(0) = \color{red}{1}
\end{aligned}
$$
Solved with Taylor expansion
A: As Salahamam_Fatima answered, using equivalents makes the problem of the limit quite simple.
You could even go further and see how is approached the limit
$$A=\left[1 + x\sin(\pi - x) \right]^{1/x}=\left[1 + x\sin( x) \right]^{1/x}$$ Take logarithms $$\log(A)=\frac 1x \log\left[1 + x\sin( x) \right]$$ Now, use Taylor expansion $$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$ $$\log(A)=\frac 1x \log\left[1+x^2-\frac{x^4}{6}+O\left(x^5\right) \right]$$ Now, use the expansion of $\log(1+t)$ and replace $t$ by $x^2-\frac{x^4}{6}+\cdots$ to get $$\log(A)=\frac 1x\left[x^2-\frac{2 x^4}{3}+O\left(x^5\right)\right]=x-\frac{2 x^3}{3}+O\left(x^4\right)$$ Now, Taylor again since $$A=e^{\log(A)}=1+x+\frac{x^2}{2}-\frac{x^3}{2}+O\left(x^4\right)$$ which, for sure, shows the limit and how it is approached.
Just for curiosity, plot on the same graph the original function and the above approximation. You will probably be surprised to see how close they are up to $x=\frac 12$.
For the fun, using your pocket calculator, compute the value of $A$ when $x=\frac \pi 6$. The exact value would be $$A=\left(1+\frac{\pi }{12}\right)^{6/\pi }\approx 1.5591$$ while the approximation would give $$A=1+\frac{\pi }{6}+\frac{\pi ^2}{72}-\frac{\pi ^3}{432}+\cdots \approx 1.5889$$
To end, suppose that you need to solve rigorously for $x$ the equation $$\left[1 + x\sin(\pi - x) \right]^{1/x}=\frac 32$$ and that the numerical method (Newton for example) requires a "reasonable" guess, ignore the cubic term and solve the quadratic $$1+x+\frac{x^2}{2}=\frac 32\implies x=\sqrt 2-1\approx 0.4142$$ while the exact solution would be $\approx 0.4622$. Newton iterates would then be
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 0.414214 \\
 1 & 0.461216 \\
 2 & 0.462236 \\
 3 & 0.462236
\end{array}
\right)$$
A: The answer to the more general question "How do we find $\lim_{x \to a}\{f(x)\}^{g(x)}$?" is that we take logarithms. Thus if $L$ is the desired limit then
\begin{align}
\log L &= \log\left(\lim_{x \to 0}(1 + x\sin (\pi - x))^{1/x}\right)\notag\\
&= \lim_{x \to 0}\log(1 + x\sin (\pi - x))^{1/x}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0}\frac{\log(1 + x\sin x)}{x}\notag\\
&= \lim_{x \to 0}\sin x \cdot\frac{\log(1 + x\sin x)}{x\sin x}\notag\\
&= 0 \cdot 1 = 0\notag
\end{align}
and hence $L = 1$.
