Limit

Calculus Level pending

$\large \lim_{x\to0^+} \dfrac{ \displaystyle \int_0^{\arctan x} \sin(t^2) \, dt}{x \cos x - x}$

Find the value of the closed form of the above limit to 3 decimal places.

The answer is -0.667.

1 solution

Joao Miguel Coelho
Jan 25, 2017

It's known that the undefined integral of a integrable function is a continous function. So there's an indetermination, $\frac{0}{0}$ .

Applying L'Hopital:

$\huge\lim_{x \rightarrow 0^{+}}\frac{(\int_{0}^{arctan(x)}sin(t^2)dt)^{'}}{(xcos(x)-x)^{'}} = {\huge\lim_{x \rightarrow 0^{+}}\frac{ {sin(arctan^2(x))\frac{1}{1+x^2}}}{cos(x)-xsin(x)-1}}$

${\huge=\lim_{x \rightarrow 0^{+}}\frac{{sin(arctan^2(x))}}{arctan^2(x)}\times\frac{arctan^2(x)}{x^2}\times\frac{1}{1+x^2}\times\frac{x^2}{cos(x)-xsin(x)-1}}$

${\huge=\lim_{x \rightarrow 0^+}\frac{x^2}{cos(x)-xsin(x)-1}}$

Applying L'Hopital again:

${\huge=\lim_{x \rightarrow 0^+}\frac{(x^2)^{'}}{(cos(x)-xsin(x)-1)^{'}} {\huge=\lim_{x \rightarrow 0^+}\frac{2x}{-2sin(x)-xcos(x)}}}$

${\huge=\lim_{x \rightarrow 0^+}\frac{2}{-2\frac{sin(x)}{x}-cos(x)}=-\frac{2}{3}\approx-0.667}$

$(\int_{0}^{arctan(x)}sin(t^2)dt)^{'}$

Let $f(x)=\int_{0}^{x}sin(t^2)dt$ and $g(x)=arctan(x)$

$f(g(x))=\int_{0}^{arctan(x)}sin(t^2)dt$

$f(g(x))^{'}=f^{'}(g(x))\times g^{'}(x)$

$(\int_{0}^{arctan(x)}sin(t^2)dt)^{'}=f^{'}(g(x))\times g^{'}(x)=sin(arctan^2(x))\times\frac{1}{1+x^2}$

What is sen(x).

Can u xplain the second step ?

Kushal Bose - 4 years, 4 months ago

Sen(x) is how I write sin(x), i try to avoid it here but sometimes it happens, already edited, thanks.

Now the second step:

$\frac{ {sin(arctan^2(x))\frac{1}{1+x^2}}}{cos(x)-x-xsin(x)-1}=\frac{ {sin(arctan^2(x))}}{(cos(x)-x-xsin(x)-1)(1+x^2)}=\frac{ {sin(arctan^2(x))(arctan^2(x))}}{(cos(x)-x-xsin(x)-1)(1+x^2)(arctan^2(x))}$

As $x \rightarrow 0 , arctan^2(x) \rightarrow 0$ , so $\frac{sin(arctan^2(x))}{arctan^2(x)}=1$

We are left with $\frac{ {(arctan^2(x))}}{(cos(x)-x-xsin(x)-1)(1+x^2)}$ and it's known that $\frac{arctan(x)}{x}=1$ as $x \rightarrow 0$ :

$\frac{ {(arctan^2(x))x^2}}{(cos(x)-x-xsin(x)-1)(1+x^2)x^2}=\frac{1}{1+x^2}(\frac{arctan(x)}{x})^2\frac{x^2}{cos(x)-x-xsin(x)-1}$

When taking limits they'll all be one exept for the third fraction.

(I did this on my mobile phone, when i get home if you can't understand i'll do it more clearly on my computer)

Joao Miguel Coelho - 4 years, 4 months ago

Differentiation of $xcosx-x$ will be $cosx-x sinx -1$ but you have written something other.Can u recheck that ?

Kushal Bose - 4 years, 4 months ago

@Kushal Bose – Yes you are right, $-x$ was wrong there, but was just a typo because as you can see i didnt use it anywhere else

Joao Miguel Coelho - 4 years, 4 months ago

Limit

The answer is -0.667.

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