Here is the problem.

Let $f$ be the function that has the value of $f(1)=1$ and $f'(1)=2$. Find the value of $$ L = \lim_{x \to 1} {\frac{\arctan{\sqrt{f(x)}-\arctan{f(x)}}}{ \left (\arcsin{\sqrt{f(x)}}-\arcsin{f(x)}\right)^2}} $$

I have tried using $$ L=\lim_{x\to 1} \frac{1}{x-1}\frac{\frac{\arctan{\sqrt{f(x)}}-\arctan{\sqrt{f(1)}}}{x-1}-\frac{\arctan{{f(x)}}-\arctan{{f(1)}}}{x-1}} {\left [\frac{\arcsin{\sqrt{f(x)}}-\arcsin{\sqrt{f(1)}}}{x-1}-\frac{\arcsin{{f(x)}}-\arcsin{{f(1)}}}{x-1} \right ]^2} $$ and reduced that big chunks by using $f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$ which I got $$ \begin{split} L&=\lim_{a\to 1} \frac{1}{a-1}\frac{\Big [ \arctan\sqrt{f(x)} \Big ]'_{x=a} - \Big [ \arctan{f(x)} \Big ]'_{x=a}} {\Big [ \arcsin\sqrt{f(x)} \Big ]'_{x=a} - \Big [ \arcsin{f(x)} \Big ]'_{x=a}}\\[2em] &=\lim_{a\to 1} \frac{1}{a-1} \frac{\frac{1}{1+f(a)}\frac{1}{2\sqrt{f(a)}}f'(a)-\frac{1}{1+(f(a))^2}f'(a)} {\left [ \frac{1}{\sqrt{1-f(a)}}\frac{1}{2\sqrt{f(a)}}f'(a)-\frac{1}{\sqrt{1-(f(a))^2}}f'(a) \right ]^2}\\[2em] &=\lim_{a\to 1} \frac{1}{f'(a)\frac{a-1}{1-f(a)}} \frac{\frac{1}{1+f(a)}\frac{1}{2\sqrt{f(a)}}-\frac{1}{1+(f(a))^2}} {\left [ \frac{1}{2\sqrt{f(a)}}-\frac{1}{\sqrt{1+f(a)}} \right ]^2}\\[2em] &=\lim_{a\to 1}{-\frac{\frac{1}{1+f(a)}\frac{1}{2\sqrt{f(a)}}-\frac{1}{1+(f(a))^2}} {\left [ \frac{1}{2\sqrt{f(a)}}-\frac{1}{\sqrt{1+f(a)}} \right ]^2}}\\[2em] &=\boxed{(\sqrt{2}+1)^2} \end{split} $$ but the answer keys tell me that the answer of this problem is $L=\left( \frac{\sqrt{2}+1}{2}\right)^2$. So, Can someone please explain to me what did I do wrong?

  • 1
    $\begingroup$ +1 for all the efforts you made $\endgroup$ – Deepesh Meena Sep 14 '18 at 13:25
  • $\begingroup$ You could plug in f (x)=1+2(x-1) then use Wolfram Alpha to see if your answer is correct. $\endgroup$ – ericf Sep 14 '18 at 14:27

There are several steps that are not clearly allowed.

The first step is okay: $$ \frac{ \arctan \sqrt{f(x)} - \arctan f(x) }{ \left( \arcsin \sqrt{f(x)} - \arcsin f(x) \right)^2 } = \frac{ (\arctan \sqrt{f(x)} - \arctan 1) - (\arctan f(x) - \arctan 1) }{ \left( (\arcsin \sqrt{f(x)} - \arcsin 1) - (\arcsin f(x) - \arcsin 1) \right)^2 } \\ = \frac{1}{x-1} \frac{ \frac{\arctan \sqrt{f(x)} - \arctan 1}{x-1} - \frac{\arctan f(x) - \arctan 1}{x-1} }{ \left( \frac{\arcsin \sqrt{f(x)} - \arcsin 1}{x-1} - \frac{\arcsin f(x) - \arcsin 1}{x-1} \right)^2 } \\ $$

Then you assume not only that the limit can be distributed over the internal terms, which would give derivatives at $x=1$, e.g. $\left[ \arctan\sqrt{f(x)} \right]'_{x=1},$ but even that you can take the limits of the derivatives, $\lim_{a \to 0} \left[ \arctan\sqrt{f(x)} \right]'_{x=a}.$

Also, as gimusi has already pointed out, $\arcsin \sqrt{f(x)}$ and $\arcsin f(x)$ are not differentiable when $f(x)=1.$

  • $\begingroup$ Thanks for the quote! $\endgroup$ – gimusi Sep 14 '18 at 23:36

Probably the derivation is wrong because arcsin x is not differentiable at $x=1$, therefore your first step is not allowed.

As an alternative, since $f(x)$ is continuos at $x=1$ we have that

$$L = \lim_{y \to 1^-} {\frac{\arctan{\sqrt{y}-\arctan{y}}}{ \left (\arcsin{\sqrt{y}}-\arcsin{y}\right)^2}}$$

then we can use that

  • $\arcsin(x) = \arctan\left(\frac{x}{\sqrt{1 - x^2}}\right)$
  • $\arctan(u) - \arctan(v) = \arctan\left(\frac{u - v}{1 + uv}\right) $

to obtain

$$L = \lim_{y \to 1^-} \frac{ \arctan\left( \frac{\sqrt y -y}{1+y\sqrt y}\right)} { \left[\arctan \left(\frac{\sqrt y \sqrt{1-y^2}-y\sqrt{1-y}} { \sqrt{(1-y)(1-y^2)}+y\sqrt y} \right)\right]^2 }=\frac{3+ 2\sqrt 2}{4}=\left( \frac{\sqrt{2}+1}{2}\right)^2$$

indeed by $y=1-u$ with $u\to 0^+$ by binomial expansion $\sqrt y = 1-\frac12 u +o(u)$ it easy to show that

  • $\arctan\left( \frac{\sqrt y -y}{1+y\sqrt y}\right)\sim \frac 14 u$
  • $\arctan \left(\frac{\sqrt y \sqrt{1-y^2}-y\sqrt{1-y}} { \sqrt{(1-y)(1-y^2)}+y\sqrt y} \right)\sim (\sqrt 2 -1) \sqrt u$

and therefore

$$\frac{ \arctan\left( \frac{\sqrt y -y}{1+y\sqrt y}\right)} { \left[\arctan \left(\frac{\sqrt y \sqrt{1-y^2}-y\sqrt{1-y}} { \sqrt{(1-y)(1-y^2)}+y\sqrt y} \right)\right]^2 }\sim \frac{\frac 14 u}{(\sqrt 2 -1)^2 u}=\frac{3+ 2\sqrt 2}{4}$$

  • $\begingroup$ I think it is "differentiable" instead of "derivable". But good answer nevertheless! (+1) $\endgroup$ – Math_QED Sep 15 '18 at 11:05
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    $\begingroup$ Thanks I fix that! Bye $\endgroup$ – gimusi Sep 15 '18 at 11:06

We shall need several times the following principle: If $f(0)=0$, $\>f'(0)=1$, and $\lim_{x\to0} g(x)=0$, then we may write $$f\bigl(g(x)\bigr)=g(x)\>h(x)\ ,\tag{1}$$ whereby the function $x\mapsto h(x)$ defined by $(1)$ satisfies $\lim_{x\to0}h(x)=1$.

We only need that $f(x)\nearrow1$ when $x\nearrow 1$ (note that your expression is undefined when $x>1$). There are functions $x\mapsto \alpha:=\alpha(x)$ and $x\mapsto\beta:=\beta(x)$ with $$0<\alpha<\beta\>,\qquad \lim_{x\to1-}\alpha(x)=\lim_{x\to1-}\beta(x)=0\ ,$$ such that $$\sqrt{f(x)}=\cos\alpha\>,\qquad f(x)=\cos\beta\ .$$ This ensures $$\arcsin\sqrt{f(x)}-\arcsin f(x)=\beta-\alpha\ .$$ On the other hand $$\arctan\sqrt{f(x)}-\arctan f(x)=\arctan{\cos\alpha-\cos\beta\over 1+\cos\alpha\cos\beta}={1\over2}(\cos\alpha-\cos\beta)\> g(x)\>,$$ whereby $\lim_{x\to1-}g(x)=1$. Now $${1\over2}(\cos\alpha-\cos\beta)=\sin{\beta+\alpha\over2}\sin{\beta-\alpha\over2}={1\over4}(\beta+\alpha)(\beta-\alpha) h(x)\ ,$$ whereby $\lim_{x\to1-} h(x)=1$. It follows that $$L={1\over4}\lim_{x\to1-}{\beta+\alpha\over\beta-\alpha }={1\over4}\lim_{x\to1-}{1+{\alpha\over\beta}\over1-{\alpha\over\beta}}.$$ As $\cos^2\alpha=\cos\beta$ for all $x<1$ we have $\>\sin^2\alpha=2\sin^2{\beta\over2}$, which then leads to $\lim_{x\to1-}{\alpha\over\beta}={1\over\sqrt{2}}$, so that we end up with $$L={1\over4}\bigl(3+2\sqrt{2}\bigr)\ .$$


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