# $\lim _{x\to 0}\left(\frac{\tan\left(x\right)-x}{x-\sin\left(x\right)}\right)$ without L'Hopital's Rule

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.

• 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. – The Count Jan 31 '17 at 18:15
• @TheCount Are you sure only that works for the first one? Because I get an indetermined form when using that... – DonAntonio Jan 31 '17 at 18:23
• @TheCount: How exactly are you using that in the first case? – Henning Makholm Jan 31 '17 at 18:23
• @DonAntonio Well, it sure looks like one that it would work on. I'll write it out to check. – The Count Jan 31 '17 at 18:25
• @HenningMakholm see above (and soon below) – The Count Jan 31 '17 at 18:25

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

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

• Always a topic of discussion: If L'Hospital rule is not allowed (because it involves derivatives), are Taylor expansions then allowed? – imranfat Jan 31 '17 at 18:27
• 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. – DonAntonio Jan 31 '17 at 18:28
• 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. – Henning Makholm Jan 31 '17 at 18:36
• 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. – Vincenzo Oliva Jan 31 '17 at 18:37
• 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. – DonAntonio Jan 31 '17 at 18:49