I'm stuck after i tried to integrate by parts.
$\displaystyle \lim x \int_x^1 \frac{cos t}{t^2} dt = 1$
The limit is actually as $\displaystyle x \rightarrow 0^+$ , I can't seemed to key in the input.
Thanks in advance.
I wonder if the comparison test would work, which you find a function similar to
the function which in this case you can do, someone correct me if I am wrong
since cosine is bounded by -1 and 1
$\displaystyle -1\le cos(t)\le1$
so then
$\displaystyle \frac{-1}{t^2}\le \frac{cos(t)}{t^2}\le \frac{1}{t^2}$
so in comparsion, you can say you function behaves like the function
$\displaystyle \frac{1}{t^2}$, if you dont know the comparsion test, look it up it can prove helpful, if you find a function larger than your original then you the function you were given which is small will also converge, if you find a function smaller than the original diverges then the original will diverge, this does not apply to vice versa so now you can do
$\displaystyle \lim_{x\to0+}\int_x^1\frac{1}{t^2}$
$\displaystyle \lim_{x\to0+}-\frac{1}{t}\|_x^1$
you should be able to take it from here
Is this $\displaystyle \lim_{x\to0^+}x\int_x^1\frac{\cos(z)}{z^2}dz$ or $\displaystyle \lim_{x\to0^+}\int_x^1\frac{\cos(z)}{z^2}dz$
Seeing your asnwer I assume it's the first.
I hate using it, but I assume it's how this is to be done
$\displaystyle \lim_{x\to0^+}\frac{\int_x^1\frac{\cos(z)}{z^2}dz} {\frac{1}{x}}$
Both the denominator and numerator approach infinity as $\displaystyle x\to0^+$. Apply L'hopital's rule and...done.
You would need to squeeze it on both sides, not just one.
$\displaystyle \lim_{x\to0^+}x\int_x^1-\frac{1}{t^2}\,dt\leq\lim_{x\to0^+}x\int_x^1\frac{ \cos t}{t^2}\,dt\leq\lim_{x\to0^+}x\int_x^1\frac{1}{t^2 }\,dt$
So $\displaystyle \lim_{x\to0^+}x\cdot\frac{x-1}{x}\leq\lim_{x\to0^+}x\int_x^1\frac{\cos t}{t^2}\,dt\leq\lim_{x\to0^+}x\cdot\frac{1-x}{x}$
So all this tells you is that $\displaystyle -1\leq\lim_{x\to0^+}x\int_x^1\frac{\cos t}{t^2}\,dt\leq1$
Or maybe since it goes to zero from the positive side, $\displaystyle 0\leq\lim_{x\to0^+}x\int_x^1\frac{\cos t}{t^2}\,dt\leq1$