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I struggle with solving this with L'Hospital technique:

$$\lim_{x\to0} \frac{ax-\arctan(ax)}{bx-\arctan(bx)} \tag{1}$$

It should be:

$$\lim_{x\to0} \dfrac{a-\dfrac{a}{a^2x^2+1}}{b-\dfrac{b}{b^2x^2+1}} \tag{2}$$

I end up with:

$$\frac{a-a/1}{b-b/1} \tag{3}$$

Which is $0/0$.

Just the same as if I didn't use l'Hospital method.

Same if "tan" is used instead of "arctan".

Yet, it seems it is solvable.

Any clue in order to get me on the right path?

Thanks.

Blue
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Bachir Messaouri
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4 Answers4

3

Hint:

Use

$$a-\dfrac a{a^2x^2+1}=\dfrac{a^3x^2}{a^2x^2+1}$$

Alternatively

Using Are all limits solvable without L'Hôpital Rule or Series Expansion

$$\lim_{x\to0}\dfrac{ax-\arctan(ax)}{(ax)^3}=\dfrac13$$

lab bhattacharjee
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  • Thanks. Doesn't your numerator still produce a zero when applying the limit? PS: I don't see anything that resemble to my case in your link and I don't understand how the last line produces a 1 on the numerator. Any further help would be appreciated. – Bachir Messaouri Sep 27 '19 at 10:11
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    @BachirMessaouri, By your method $$\lim_{x\to}\dfrac{a^3x^3}{b^3x^3}=?$$ – lab bhattacharjee Sep 27 '19 at 10:14
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    @BachirMessaouri, Using my link, $$\dfrac{ax-\arctan(ax)}{bx-\arctan(bx)}=\dfrac{a^3}{b^3}\cdot\dfrac{\dfrac{ax-\arctan(ax)}{(ax)^3}}{\dfrac{bx-\arctan(bx)}{(bx)^3}}$$ – lab bhattacharjee Sep 27 '19 at 10:15
  • I spoke too soon. I still don't see it. How are you having a^3x^3/b^3x^3 from your line above (with a^3x^2)? I'm lost. – Bachir Messaouri Sep 27 '19 at 10:20
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    @BachirMessaouri, Do you know $$\lim_{h\to0}\dfrac hh=1?$$ – lab bhattacharjee Sep 27 '19 at 10:21
  • I see that you can simplify x^3/x^3 but how did you get there from a^3x^2/a^2x^2+1 (sorry, I really try)? – Bachir Messaouri Sep 27 '19 at 10:23
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    @BachirMessaouri $$a-\dfrac a{a^2x^2+1}=\dfrac{a(a^2x^2+1)-a}{a^2x^2+1}=?$$ – lab bhattacharjee Sep 27 '19 at 10:27
  • This one I got. it's the transition from "a^3x^2/a^2x^2+1" to" ^3^3" that I don't (thank you for your patience) – Bachir Messaouri Sep 27 '19 at 10:29
  • @BachirMessaouri, Take out $$\lim_{x\to0}\dfrac{b^2x^2+1}{a^2x^2+1}$$ – lab bhattacharjee Sep 27 '19 at 10:32
  • I start to see where you are going and why you end up with a^3/b^3 by simplifying. What's beyond me is why you would take those two out (in your last comment) as what you are taking out is mixing b^2 and a^2. How in the world can you simplify "b^2x^2+1/a^2x^2+1....." and get 1/1?. Obviously it's there in front of my face and I don't see it. – Bachir Messaouri Sep 27 '19 at 10:39
  • Ok, this charade thing, piece by piece got me confused. Now I see it. You cancel out the extra pieces AFTER applying the limit which leaves you with the a^3/b^3. I was understanding that you made the simplification before applying the limit which got me very confused. Thank you for your help. – Bachir Messaouri Sep 27 '19 at 10:51
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$\lim_{x \to 0} \frac {ax - \arctan (ax)} {bx - \arctan (bx)}=\lim_{x \to 0} \frac {a-a/(1+a^{2}x^{2})} {b-b/(1+b^{2}x^{2})}$. Simplify this ratio. The limit is $\frac {a^{3}} {b^{3}}$ since the ratio simplifies to $\frac {a^{3}} {b^{3}} \frac {1+b^{2}x^{2}} {1+a^{2}x^{2}}$ after cancelling $x^{2}$.

Kavi Rama Murthy
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  • Thank you very much. I hesitated to consider your answer as the one (because it is), but lab bhattacharjee made the effort to tell the same thing you did repeatedly. I just didn't get it right away. Thanks! – Bachir Messaouri Sep 27 '19 at 10:55
  • @BachirMessaouri My intention was only to point out that instead of directly putting $x=0$ after one application of L'Hopital's Rule you could have done some simplification to get the answer. My comment has nothing to do with voting or approval. I think you were right in approving Lab Bhattacarjee's answer. – Kavi Rama Murthy Sep 27 '19 at 11:52
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We have

$$ \lim_{x\to0} \dfrac{ax-\arctan(ax)}{bx-\arctan(bx)} = \lim_{x\to0}\frac{\sin\left(ax-\arctan(ax)\right)}{\sin\left(bx-\arctan(bx)\right)} = \lim_{x\to 0}\frac{\sin(a x)-a x \cos(ax)}{\sin(b x)-a x \cos(b x)}\frac{\sqrt{1+b^2x^2}}{\sqrt{1+a^2x^2}} $$

now discarding $\frac{\sqrt{1+b^2x^2}}{\sqrt{1+a^2x^2}}$ we follow with

$$ \lim_{x\to 0}\frac{\sin(a x)-a x \cos(ax)}{\sin(b x)-a x \cos(b x)} $$

Applying l'Hopital twice now, we obtain easily as result

$$ \frac{a^3}{b^3} $$

Cesareo
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You just keep on applying L'Hopital's rule. After three times you will get$$\lim_{x\to0}\frac{ax-\arctan(ax)}{bx-\arctan(bx)}=\frac{2a^3}{2b^3}=\left(\frac ab\right)^3.$$

José Carlos Santos
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