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Here are the functions:

a) $\displaystyle\lim _{x\to 0}\left(\frac{\tan\left(x\right)-x}{x-\sin\left(x\right)}\right)$

If I used L'Hopital's rule the limit is $2$

b) $\displaystyle\lim _{x\to 0}\:\frac{e^x\cdot \:\sin\left(x\right)-x\cdot \left(1+x\right)}{x^3}$

here $\dfrac{1}{3}$

c) $\displaystyle\lim _{x\to 0}\left(\frac{\ln\left(\sin\left(3 x\right)\right)}{\ln\left(\sin\left(7x\right)\right)}\right)$

and here $1$

but the problem is that I am not allowed to use L'Hopital's rule, can you give me ideas for another type of approaches?

UPDATE:

I apologize, I see there is some discussion and confusion among people, which obviously goes beyond my functions, but still I wanted to explain that I have been missing a lots of lectures recently due to illness and last week I got $0$ points for using L'hopital because we have not learnt it, so my guess was that we are not allowed this time either, but I just talked to my tutor and he told me that just in the last lecture, they introduced L'hopital rule to us so I am free to use it. I'm very sorry.

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    Hint: Use the identity $\lim\limits_{x\to 0}\frac{\sin x}{x}=1$. This will work for the first one, and I'd bet the others as well. – 2017-01-31
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    @TheCount Are you sure *only* that works for the first one? Because I get an indetermined form when using that... – 2017-01-31
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    @TheCount: How exactly are you using that in the first case? – 2017-01-31
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    @DonAntonio Well, it sure looks like one that it would work on. I'll write it out to check. – 2017-01-31
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    @HenningMakholm see above (and soon below) – 2017-01-31
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    @DonAntonio You are completely correct. Thank you for catching that. – 2017-01-31
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    This is a clever question. Fooled me! – 2017-01-31
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    Should I delete the original comment, or leave it there for context? – 2017-01-31
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    Hi I read your question and some of the comments, I would to know what techniques you are expected to use ? I ask because $\lim \frac{\sin x -x}{x^3}$, which is what the first limit boils down to is very hard without Taylor or lHospital. – 2017-01-31
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    @ReneSchipperus Hi, I just updated my question. – 2017-01-31
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    No worries, without l'Hospital questions are very popular at the moment, I like them. – 2017-01-31

1 Answers 1

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a) use the fact that $\tan{x}=x+x^3/3+o(x^3)$ and $\sin{x}=x-x^3/6+o(x^3)$ to get

$${\tan{x}-x\over x-\sin{x}}={{x^3\over 3}+o(x^3)\over {x^3\over 6}+o(x^3)}=2+o(1)$$

So the limit is $2$

b) for this one we need $e^x=1+x+x^2/2+o(x^2)$ to write

$${e^x\cdot\sin{x}-x(1+x)\over x^3}={x+x^2+{x^3\over 3}-x-x^2+o(x^3)\over x^3}={1\over 3}+o(1)$$

So the limit is $1/3$

c) the last one is even simpler

$${\ln{\sin{7x}}\over \ln{\sin{3x}}}={\ln{7x}+\ln{\sin{7x}\over 7x}\over \ln{3x}+\ln{\sin{3x}\over 3x}}$$

Now keeping in mind $\sin{x}/x\to 1$ the limit is equal to

$$\lim_{x\to 0}{\ln{7}+\ln{x}\over \ln{3}+\ln{x}}=\lim_{x\to 0}{{\ln{7}\over \ln{x}}+1\over {\ln{3}\over \ln{x}}+1}=1$$

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    Always a topic of discussion: If L'Hospital rule is not allowed (because it involves derivatives), are Taylor expansions then allowed? – 2017-01-31
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    @imranfat my instinct is to say yes, since they are a more basic application, but this is always a good point. can you think of another way to do it? – 2017-01-31
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    Good point by imranfat. Of course this could be allowed, as a means to *force* students to use Taylor developments, yet without any explantion it'd be absurd, at least imo, to allow Taylor but forbidding l'Hospital. – 2017-01-31
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    :D you beat me to this! (I can't type on phone very well) – 2017-01-31
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    Especially since what you _actually do_ when finding and using the Taylor expansions is exactly the same as using L'Hospital repeatedly until you reach nonzero coefficients -- just with some identical factorials above and below, and with some boilerplate arguments about why the procedure works added in. – 2017-01-31
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    I'll quote myself: Taylor expansions and L'Hospital rule are essentially equivalent, even if not necessarily in practice. I think no-L'Hospital (and no-Taylor) exercises are a good tool to develop creativity and reasoning skills of students. – 2017-01-31
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    @imranfat please see here: http://meta.math.stackexchange.com/questions/18857/using-taylor-expansion-on-a-limit-tagged-without-lhospital – 2017-01-31
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    In defense of series expansion: they work better than L'Hospitals most of the time, you can **define** these functions via their Taylor series, etc. – 2017-01-31
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    @SimplyBeautifulArt: Sure. But you don't, in high school. – 2017-01-31
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    And how should that affect me? @VincenzoOlivia (I am in high school...) – 2017-01-31
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    @SimplyBeautifulArt: Just a critique to your previous comment, in the context of these exercises – 2017-01-31
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    @VincenzoOliva Please, try computing $\lim_{x\to0}\frac{\tan x-x}{x^3}=\frac{1}{3}$, without Taylor or l'Hôpital. – 2017-01-31
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    I dont think there's a l'Hospital vs Taylor expansions discussion here. To each its own. It is the logic behind the use of Taylor expansions to evaluate limits **when l'Hospital** is forbidden that we're debating (or sort of...). I commented above one possible reason, but in this case that seems to be exaggerated. – 2017-01-31
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    @VincenzoOlivia Sure, but you could've easily underestimated the OP's age as well as a certain high schooler's mathematical knowledge. – 2017-01-31
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    I seriously doubt there are many high schools around the globe that deal with the kind of limits this question asks, and even worse: requiring the limit *not* be evaluated with l'Hospital. – 2017-01-31
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    @DonAntonio I know a 15 year old in real analysis. – 2017-01-31
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    @SimplyBeautifulArt and I know a professor (a woman, in fact) who got her B.Sc. at 15 and her PhD at 19 (with no less than Atiyah in the UK...) and it is now a professor in the Hebrew University in Jerusalem. *Still*, I think I am not exaggerating when I say **most** people in the world don't *usually* do those things at those ages... – 2017-01-31
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    :-) and now I feel old and that i should've started studying math at age 10. – 2017-01-31
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    @egreg: That's doable, if we assume its existence. It resembles something I've already done on MSE. It equals $$\ell=\lim_{x\to0}\frac{\sin x -x\cos x}{x^3}=\lim_{x\to0}\frac{\sin x -x}{x^3}+\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12+\lim_{x\to0}\frac{\sin x -x}{x^3}.$$ Then we try to get information from a very similar limit: $$\lim_{x\to0}\frac{\sin x\cos x -x}{x^3}=\lim_{x\to0}\frac{\sin(2x)/2 -x}{x^3}=\lim_{x\to0}\frac{\sin(2x) -2x}{2x^3} =\lim_{x\to0}\frac{\sin x -x}{\frac14 x^3}=4(\ell-\frac12)=4\ell-2.$$ ... – 2017-01-31
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    ... Therefore $$3\ell-2=\lim_{x\to0}\frac{\sin x\cos x +x\cos x -\sin x -x}{x^3}=\lim_{x\to0}\frac{\sin x+x}{x}\frac{\cos x-1}{x^2}=-1,$$ *i.e.* $\ell=\frac13$. – 2017-01-31
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    @SimplyBeautifulArt: Excuse me? I didn't mean to offend anyone. I simply said Taylor expansions are usually covered in the first year of university. And this is why I think OP's exercise should be for highschoolers, otherwise, in my opinion, it is poorly worded and should have explicitly mentioned Taylor expansions. In my experience, undergraduates are encouraged to use Taylor explicitly – 2017-01-31
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    @VincenzoOliva I think Simply was referring to me and my past comment... – 2017-01-31
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    @VincenzoOliva do not restrict me to the "we all succeed together" program. :-) – 2017-01-31
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    @VincenzoOliva Why is it "obvious" that $$\;\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12\;??$$ – 2017-01-31
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    @DonAntonio: $$\frac{1-\cos x}{x^2}=\frac{1-\cos^2x}{x^2(1+\cos x)}\sim\frac12 \left(\frac{\sin x}{x}\right)^2$$ and $\lim\limits_{x\to0}\frac{\sin x}{x}=1$ is proved geometrically. – 2017-01-31
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    @SimplyBeautifulArt: I have said nothing about you. Bye, peace – 2017-01-31
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    @VincenzoOliva without Taylor, how do you prove that equivalence rigorously? – 2017-01-31
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    @SimplyBeautifulArt: Taylor isn't necessary to prove $1+\cos x\to 2$. – 2017-01-31
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    @VincenzoOliva Just as a data point: I learnt Taylor expansion in high school, and I've never formally learned L'Hospital's rule in either school or university. I first encountered that in an aside remark in a book (on complex analysis; the topic of the section was computing residues at poles). For me, using Taylor expansions is natural and basic, while L'Hospital's rule is exotic. – 2017-01-31
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    @DanielFischer: That's interesting to know, I guess programs heavily vary from country to country. – 2017-01-31
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    I too learned Taylor first by proving Euler'formula for complex exponents and complex analysis, where L'Hospitals is just truly unnecessary. – 2017-01-31
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    @VincenzoOliva or its not the program, but the learner. Likewise, learning from books is different than courses, and courses vary themselves. – 2017-01-31
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    @VincenzoOliva Yes. Things also vary in time. By the time my nieces got into high school, induction was no longer in the curriculum. I'm not sure about Taylor expansions, but I wouldn't be surprised if they are no longer taught in school here any more. – 2017-01-31
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    @DanielFischer: I see. However, I still abide by what I wrote in my first comment. Taylor expansions are very natural and useful, but if a limit is not too far in a Taylor series, not using them might push a student's thinking. Anyway, this is just my opinion. – 2017-01-31
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    @SimplyBeautifulArt: I've been talking about programs. Now, comments are not for conversations, I hope everything is fine. – 2017-01-31
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    @VincenzoOliva That's cheating: you can't assume the limit exists. Good luck. – 2017-01-31
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    @egreg: Perhaps, after all you make the rules. But the kind of exercises I've been speaking of do assume the existence, demanding clever manipulation. There are also cases when one gets away with comparison or other things, so also proving the existence, but that's a different matter. – 2017-01-31
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    That's quite a lot of discussion. I want to highlight another issue here. Ignoring terms like $\log((\sin 7x)/7x)$ should be justified properly by dividing numerator and denominator by $\log x$. One can't simply replace sub expressions by their limits while evaluating limit of a bigger expression. – 2017-01-31