I am having trouble finding the limit for the following function:     $\lim_{x \to 0} f(x)=\frac{\tan 3x}{\tan 2x}$   I do not want to use L'Hospital's rule.  I know that $\frac{\tan 3x}{\tan 2x} = \frac{\sin 3x \cos 2x}{\sin 2x \cos 3x}$, but that does not help me.  Can you maybe give me a hint? Thanks.
120943
$\lim_{x\to 0}\frac{\tan3x}{\sin2x}$= $\lim_{x\to 0}\frac{\frac{\sin(3x)}{\cos(3x)}}{\sin2x}=\lim_{x\to 0}\frac{\sin3x}{1}\cdot\frac{1}{\cos(3x)}\cdot\frac{1}{\sin(2x)}$  From this point I am lost. I believe I can pull a 3 and 2 out but I am not sure how. Can someone give me detailed instructions for a person in Calculus 1?
70158
Find $$\lim_{x \to 0} (\frac{\tan(x)}{x})^{\frac{1}{x}}$$   My first idea to solve this was to try to evaluate it and then apply the L'Hospital's rule. This is what I managed to achieve: $$\left(\frac{\tan(x)}{x}\right)^\frac{1}{x}=e^{\frac{1}{x}\ln\left(\frac {\tan(x)}x \right)} $$ However, this is problematic, because the L'Hospital's rule is not applicable in the exponent, therefore my transformation was not a good shot. What transformation should I apply to get the desired form?
86721
Calculate $$\lim_{x\to 0}\dfrac{1}{x^{\sin(x)}}$$  I'm pretty much clueless here, only that there is L'hospital obviously here.   Would appreciate any help.
86516
Find the following limit: $$\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$$   Well I tried to do the $\exp\left(\frac{ \ln\frac{\sin x}{x}}{x^2}\right)$ then apply LHR but I seem to get to endless dervivations...   There's got to be a more simple approach.
90878
Find the limit: $$\lim_{x\to0}{\frac{x^2\cdot\sin\frac{1}{x}}{\sin x}}$$ After treating it with l'Hopital rule, we get: $$\lim_{x\to0}{\frac{2x\cdot\sin \frac{1}{x}-\cos\frac{1}{x}}{\cos x}}$$ Now, the numerator of fraction doesn't have a limit, so I can't use l'Hopital rule again. What to do? I can split it into two fractions, but I'm not sure how would it help: $$\lim_{x\to0}{\frac{2x\cdot\sin \frac{1}{x}}{\cos x}} - \lim_{x\to0}{\frac{\cos\frac{1}{x}}{\cos x}}$$
73476
Find: $$\lim\limits_{x \to 1} \left( {\frac{x}{{x - 1}} - \frac{1}{{\ln x}}} \right) $$  Without using L'Hospital or Taylor approximations  Thanks in advance
49258
The limit is   $$\lim_{x\rightarrow0} \frac{x-\sin_n(x)}{x^3},$$  where $\sin_n(x)$ is the $\sin(x)$ function composed with itself $n$ times:  $$\sin_n(x) = \sin(\sin(\dots \sin(x)))$$  For $n=1$ the limit is $\frac{1}{6}$, $n=2$, the limit is $\frac{1}{3}$ and so on.  Can we define a recurrent relation upon that given hypothesis? Also, how do I involve $n$ into calculation, because the final limit will depend on it?  Any suggestions on how to tackle this?  Thank you!
123790
I have a little problem with limit of this function:  $\lim_{x \to \infty} x^2(2017^{\frac{1}{x}} - 2017^{\frac{1}{x+1}})$  I have tried de L'Hopital rule twice, but it doesn't work. Now I have no idea how to do it.
19760
Given is the function $f(x,y)=\frac{x^3-y^3}{x^2-y^2}$ defined for $(x,y)\in\mathbb{R^2}, \, x^2\neq y^2$  I need to find the limit of $\lim_{(x,y)\rightarrow(0,0)} \frac{x^3-y^3}{x^2-y^2}$.   I have tried a few paths, $(x,mx),\, (x,mx^2), \, (x,sin(x)), \, (x,tan(x))$ but all of them yield the same result, $0$.   So I tried to prove the limit exists using the epsilon-delta definition. So: $$\left|\frac{x^3-y^3}{x^2-y^2} - 0\right| = \left|\frac{(x-y)(x^2+y^2+xy)}{(x-y)(x+y)}\right|= \frac{x^2+y^2+xy}{\left|x+y\right|}&lt;\epsilon$$ Here I am stuck and do not know how to proceed.
71280
I was given a task to find  $$\lim_{x\to0}\frac{x-\sin{x}}{x^3}$$ at my school today. I thought it was an easy problem and started differentiating denominator and numerator to calculate the limit but the teacher then said we aren't allowed to use  L'Hopital's rule, but to "play around" with known limits and limit definition. I got stuck here since I can't really think of a way to do this, and according to my teacher, there are at least 4 ways. A subtle hint would be enough. Thanks.
122021
Does the limit $\lim _{x\to 0} \frac 1x \int_0^x \left|\cos \frac 1t \right| dt$ exists ? If it does then what is the value ?   I don't think even L'Hospital's rule can be applied . Please help . Thanks in advance
31079
I wonder how to do this in different way from L'Hôpital's rule:  $$\lim_{x\to 0}\frac{2\sin x-\sin 2x}{x-\sin x}.$$  Please help me solve this without using L'Hopital's rule.
94214
I was solving limits and it said $$\lim_{x\to 0} \left[\frac{x^2}{\tan(x)\sin(x)}\right]=0$$  I tried thinking about it but since I am not really good at math I couldn't proceed.  I know that $\lim_{x\to 0} \left[\frac{\sin(x)}{x}\right] = 0$ and $\lim_{x\to 0} \left[\frac{\tan(x)}{x}\right] = 1$
96491
I have a strong feeling that the following limit is zero, can anybody help me prove it.    $ \lim\limits_{(x,y,z) \to (0,0,0)} \frac{{xyz}}{{x+y+z}}$   Thanks!
27207
$$\lim _{x \rightarrow 0} \left(\frac{ \sin x}{x}\right)^{1/x}$$  I have spent an hour on the above limit and have no work to show. I tried using L'Hopital's Rule, but just kept going around in circles. Any help would be appreciated. Thank you.
38454
I am trying to evaluate the following limit without L'Hopital's:   $$\lim_{x\rightarrow 0} \frac{\sin (6x)}{\sin(2x)}$$  I know I have to use the fact that $\frac{\sin x}{x} = 1$ but I don't know how to get the limit from the above to $\frac{\sin x}{x}$ or even a portion of it to that. I know how to evaluate limits like the following   $$\lim_{x \rightarrow 0}\frac{\sin 3x}{x} = 3$$  if that's any help.  Any hints would be appreciated  Thanks!
18495
Evaluate the limit: $$ \lim_{x \to 0 } = x \cdot \sin\left(\frac{1}{x}\right) $$  So far I did:   $$ \lim_{x \to 0 } = x\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}\cdot x} $$  $$ \lim_{x \to 0 } = 1 \cdot \frac{x}{x} $$  $$ \lim_{x \to 0 } = 1 $$  Now of course I've looked around and I know I'm wrong, but I couldn't understand why. Can someone please show me how to evaluate this limit correctly? And tell me what I was doing was wrong.