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Evaluate $$\lim_\limits{x\to0^+}\frac{e^x-\cos(\lambda \sqrt x)}{\sqrt {1+\sin(\lambda x)}-1}=(*)$$

My attempt:

I have used Taylor expansion of $e^x, \ \cos x, \ \sin x:$

$$(*)=\lim_\limits{x\to0^+}\frac{1+x-1+\frac{\lambda^2 x}2+ o(x)}{\sqrt {1+\lambda x+o(x)}-1}=\lim_\limits{x\to0^+}\frac{(x+\frac{\lambda^2 x}2)(\sqrt{1+\lambda x+o(x)}+1)}{\lambda x}=\\ =\left(\frac1{\lambda}+\frac{\lambda}2\right)\lim_\limits{x\to0^+}\frac{\sqrt{1+\lambda x}+1}{\lambda x}$$

$\left(\dfrac1{\lambda}+\dfrac{\lambda}2\right)$ is the result written on my textbook, but there seems to be a typo.

Thanks in advance

P.S. the exercise comes from Calculus Problems, $8.20$ page $144$

EDIT: Actually the last step should have been: $$\left(\frac1{\lambda}+\frac{\lambda}2\right)\lim_\limits{x\to0^+}\sqrt{1+\lambda x}+1=\frac2{\lambda}+\lambda$$

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You are on the right track. At the first step, the $\lambda$ at the numerator should be squared, $$\begin{align} \lim_\limits{x\to0^+}\frac{1+x-1+\frac{\lambda^2 x}2+ o(x)}{\sqrt {1+\lambda x+o(x)}-1}&=\lim_\limits{x\to0^+}\frac{x+\frac{\lambda^2 x}2+ o(x)}{ \lambda x}\cdot \left(\sqrt{1+\lambda x+o(x)}+1\right)\\ &=\left(\frac{1}{\lambda}+\frac{\lambda }2\right)\lim_\limits{x\to0^+}(\sqrt{1+\lambda x+o(x)}+1)=\frac{2}{\lambda}+\lambda \end{align}$$ which is different from your expected result (probably a typo in your textbook).

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  • $\begingroup$ The answer given by the textbook is the one I wrote earlier in the question (I checked), but you found the value Mark Viola found. Probably my book is wrong $\endgroup$ – Lorenzo B. Mar 20 '18 at 15:30
  • $\begingroup$ Yes, it seems to be a typo. $\endgroup$ – Robert Z Mar 20 '18 at 15:32
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By standard limits

$$\frac{e^x-\cos(\lambda \sqrt x)}{\sqrt {1+\sin(\lambda x)}-1}=x\frac{\frac{e^x-1}{x}+\frac{1-\cos(\lambda \sqrt x)}{x}}{\sqrt {1+\sin(\lambda x)}-1}\frac{\sqrt {1+\sin(\lambda x)}+1}{\sqrt {1+\sin(\lambda x)}+1}=\left(\sqrt {1+\sin(\lambda x)}+1\right)\left(\frac{e^x-1}{x}+\lambda^2\frac{1-\cos(\lambda \sqrt x)}{\lambda^2x}\right)\frac{\lambda x}{\sin \lambda x}\frac1{\lambda}\\\to2\left(1+\frac12\lambda^2\right)\frac1{\lambda}=\frac2{\lambda}+\lambda$$

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  • $\begingroup$ @MarkViola Hi Mark, yes also mine! I don't know why, can we report that by flagging? $\endgroup$ – gimusi Mar 21 '18 at 20:16
  • $\begingroup$ Yes, please flag if you believe it is appropriate. $\endgroup$ – Mark Viola Mar 21 '18 at 20:22
  • $\begingroup$ @MarkViola absolutely appropriate, sorry but my +1 was already given, I can't compensate further ;) $\endgroup$ – gimusi Mar 21 '18 at 20:33
  • $\begingroup$ I wasn't fishing for an up vote, but thank you. $\endgroup$ – Mark Viola Mar 21 '18 at 22:50
  • $\begingroup$ @MarkViola Never thought about that :) $\endgroup$ – gimusi Mar 22 '18 at 2:12
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After expansion and simplification, the dominant terms of the numerator are (the units cancel out)

$$x+\frac{\lambda^2\sqrt x^2}2$$ and that of the denominator (linearize the sine, then the square root, the units also cancel out)

$$\frac{\lambda x}2.$$

Hence the ratio tends to

$$\frac2\lambda+\lambda.$$

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  • $\begingroup$ Hi Yves. Just curious ... my answer received a down vote and I'm curious whether yours did too. There are serial down voters on this site possible instances of serial down voting should be reported, I believe. ;-) $\endgroup$ – Mark Viola Mar 21 '18 at 19:48
  • $\begingroup$ @MarkViola: I don't think so. IMO, masked downvoters do not deserve that we care about them. I am more worried by the upvotes that some poor or even wrong answers get just because they were the first. $\endgroup$ – Yves Daoust Mar 21 '18 at 21:46
  • $\begingroup$ I agree with you in regards to up votes given to poor/incorrect answers. $\endgroup$ – Mark Viola Mar 21 '18 at 22:51

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