My First Nameless Problem

we have $-1\leq \sin(x) \leq 1$

so we have $-x^{n}\leq x^{n} \sin(x) \leq x^{n}$

So $\int_{0}^{n}-x^{n} dx \leq \int_{0}^{n} x^{n} \sin(x) dx \leq \int_{0}^{n}x^{n}dx$

So $\int_{0}^{n}\frac{1}{n^{n+1}}(-x^{n})dx\leq \int_{0}^{n}\frac{1}{n^{n+1}} x^{n} \sin(x)dx \leq \frac{1}{n^{n+1}} \int_{0}^{n}x^{n}dx$

So $-\frac{n^{n+1}}{(n^{n+1})(n+1)}\leq \frac{1}{n^{n+1}} \int_{0}^{n} x^{n} \sin(x)dx \leq \frac{n^{n+1}}{(n^{n+1})(n+1)}$

Taking $n\to\infty$ and using the Sandwich Rule

we have our first limit as 0 .

Now for the 2nd part...we have to evaluate the same limit as $t\to0^{+}$

we have for $0<x<1$ and $1>t>0$

$0<x^{t}< 1$

as we have to evaluate the limit as the function nears $0$

So taking an open interval $(0,\delta)$ where $0<\delta< \frac{\pi}{2}$ and let $t \in (0,\delta)$

(essentially what we want to do is we want to be only concerned about the behavior of the function with respect to $t$ in a very small deleted neighbourhood of $0$ . But since we are concerned with only the positive half of that neighbourhood it is good enough that we are working in $(0,\delta)$ , $0<\delta< \frac{\pi}{2}$ )

$0<x^{t}sin(x)<\sin(x)$ .

So $\displaystyle 0< \int_{0}^{t} \frac{1}{t^{t+1}}x^{t}\sin(x)dx < \int_{0}^{t} \frac{1}{t^{t+1}} \sin(x) dx$

which gives us:-

$0<\int_{0}^{t} \frac{1}{t^{t+1}}x^{t}\sin(x)dx < \frac{1-\cos(t)}{t\cdot t^{t}}$

We have $\displaystyle \lim_{t\to0^{+}} t^{t} = 1$ here's a good explanation video for this

and

$\displaystyle \lim_{t\to0^{+}} \frac{1-\cos(t)}{t} = \lim_{t\to0^{+}} \sin(t) = 0$ (Using L'Hopital's Rule)

So again using Sandwich Rule we have the 2nd limit as $0$

So the answer is $0+0 = 0$