Q:

Let $f(x)=\lim_{n\to \infty} (1-\sin x + e^{\frac 1n} \sin x)^n)$. If ....

Let $f(x)=\lim_{n\to \infty} (1-\sin x + e^{\frac 1n} \sin x)^n)$. If $\lim_{u\to 0} (1+u \log (1+k^2))^{\frac 1u}=2k\log^2 (f(x))$, for $k>0$ and $x\in (0,\pi)$, find $x+k$

Log is to the base $e$
$$f(x)=e^{\sin x}$$
And $$\lim_{u\to 0} (1+u\log (1+k^2))^u=e^{\log (1+k^2)}= 1+k^2$$
So $$1+k^2=2k(\sin^2x)$$
How do I find $x$ and $k$ from here?

A:

Now, $$ 0=k^2-2k\sin^2x+1=(k-1)^2+2k\cos^2x,$$ which gives $k=1$ and $x=\frac{\pi}{2}.$

