I am asked to find the following limit:  $$ \lim_{x \to 0} \frac{\tan 3x}{\tan 5x}$$  My problem is in simplifying the function. I followed two different approaches to solve  the problem. But both seems incorrect.  Apprach 1) Since $\tan \theta = \frac{sin \theta}{cos \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$, we have:  $$\frac{\tan 3x}{\tan 5x} = \frac{\sin 3x \times \cos 5x}{\sin 5x \times \cos 3x}$$  This approach does not work well, because I cannot simplify more.   Approach 2) Since $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \times \tan y)}$, we have:  $$\frac{\tan 3x}{\tan 5x} = \frac{\tan 3x}{\tan 3x + \tan 2x} = \frac{\tan 3x}{\frac{\tan 2 + \tan3}{1 - \tan 2 \times \tan 3}}$$  What am I doing wrong?
121314
I started resolving this limit:  $$\lim_{x \to 0} \dfrac{\tan 3x}{\tan 5x}$$  so:  $$\begin{align}\lim_{x \to 0} \dfrac{\tan 3x}{ \tan 5x} &amp; = \lim_{x \to 0} \dfrac{\frac{\sin 3x}{\cos 3x }}{\frac{\sin 5x}{\cos 5x}} \\ \\ &amp; = \lim_{x \to 0} \dfrac{ \sin (3x) \cdot \cos (5x)}{ \sin(5x) \cdot \cos (3x)}\\ \\ &amp; = \quad?\end{align}$$  Could someone help me please
103549
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?
22688
To find limit of $\lim_{x\to 0}\frac {\cos(\sin x) - \cos x}{x^4} $. I differentiated it using L Hospital's rule. I got $$\frac{-\sin(\sin x)\cos x + \sin x}{4x^3}\text{.}$$ I divided and multiplied by $\sin x$. Since $\lim_{x\to 0}\frac{\sin x}{x} = 1$, thus I got $\frac{1-\cos x}{4x^2}$.On applying standard limits, I get answer $\frac18$. But correct answer is $\frac16$. Please help.
99850
$$\lim_{x\to 0}\frac{10^x - 2^x - 5 ^ x + 1 } {x\tan x} $$ This is an indeterminate limit.  I want help in solving this problem. Thanks in advance
119118
I've been trying to prove the existence of this limit... trying with a bunch of different curves I'm always getting $0$ (I don't know if I'm using the suggestion correctly) but when I try to make an epsilon-delta proof I get stucked because I don't know how to properly bound the denominator.  $$\lim_{(x,y) \rightarrow (0,2)} \frac{x (y-2)^3}{3x - 5(y-2)^4}$$ (Consider curves close to $3x-5(y-2)^4=0$)  My question is: if indeed this limit does exist -and it's $0$- how can I prove it? As I wrote above I don't know how to work around the denominator. I need something smaller than $|3x-5(y-2)^4|$ ... Or maybe the limit does not exist, and then my question would be what curve can you think of to prove it. Thank you for your time!
117243
Find out the value of $a, b$ and $c$, for which $$ \lim _{x\to 0}\:\:\frac{1}{x^5}\left(\int _0^x\:\left(a+b \cos\left(t\right)+c \cos\left(2t\right)\right)dt\right)=\frac{1}{5} $$  My first thought was to separate the integral into multiple ones because i thought that it is pretty clear that each integral is a common one and a,b,c are constants and can be taken out . After I solved the integral, that's where I had the problem. I couldn't do the limit of the result that I've got in order to find out the value of $a,b$ and $c$ .
99321
Limit to evaluate: $$\lim_{x \rightarrow 0} \cfrac{\sin{(x)}+\sin{(3x)}+\sin{(5x)}}{\sin{(2x)}+\sin{(4x)}+\sin{(6x)}}$$  Proposed solution: $$ \cfrac{\sin(x)+\sin(3x)+\sin(5x)}{\sin(2x)+\sin(4x)+\sin(6x)} \Bigg/ \cdot\ \cfrac{1/x}{1/x}\Bigg/= \frac{\cfrac{\sin(x)}x + \cfrac{\sin(3x)}{3x} \cdot 3 + \cfrac{\sin(5x)}{5x} \cdot 5} {\cfrac{\sin(2x)}{2x} \cdot 2 + \cfrac{\sin(4x)}{4x} \cdot 4 + \cfrac{\sin(6x)}{6x} \cdot 6} $$  Using $\lim_{x \rightarrow 0} \frac{\sin x}x=1$, we get $$\frac{1+1\cdot 3+1\cdot 5}{1\cdot 2+1\cdot 4+1\cdot 6} = \frac 9{12} = \frac 3 4$$  Please tell me if I am correct.
8376
I need your help with evaluating this limit:  $$ \lim_{n \to \infty }\underbrace{\sin \sin \dots\sin}_{\text{$n$ compositions}}\,n,$$  i.e. we apply the $\sin$ function $n$ times.  Thank you.
25389
I have calculus homework question that to be quite frank I don't even know how to begin to solve. Here is my attempt on how to to solve the question; however I'm not sure if I'm tackling this correctly. I'm hoping that someone here can tell me if I'm doing this correctly and if my answer would be correct.  Here is the question.     Find the limit or show that it does not exist:   $$\lim_{x\to \infty}(e^{-x}+2\cos3x)$$   My attempt to the solution is DNE or does not exist simply because cosine oscillates between -1 and 1 and it never approaches any one number to reach a horizontal asymptote. Is this answer correct? Or how should I approach this problem if it's not? Or should it be  $$-\infty$$ since that would be the the value of $$e^{-x}\quad ?$$Please forgive my stupidity if I'm incorrect in all fronts.
19929
I have a problem with the calculation of the following limit. \begin{equation} \lim_{n\rightarrow \infty} \frac{1+\sqrt{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n} \end{equation} I do not know where to start! Thank you very much
118596
This was the question of a test. My question is if my attempt to solve it is correct, and if it is, why is it correct.  $$\lim_{(x,y) \to (0,0), x+y \neq 0}{\frac{\ln(1-x-y)}{x+y} } $$  My attempt:   Let $\xi = -x-y $. Then $\xi \to 0$ whenever $(x,y) \to (0,0)$ and $x+y \neq 0 \iff \xi \neq 0$. (Is it then correct to say that the previous limit exists and is equal to the following iff the following exists? And why?):  $$\lim_{\xi \to 0, \xi \neq 0}{\frac{\ln(1+\xi )}{-\xi}}$$   If it is correct, then the limit exists and is $-1$. If it is correct, why is it correct?
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: $$\lim_{x\to0}{\frac{2x\cdot\sin \frac{1}{x}}{\cos x}} - \lim_{x\to0}{\frac{\cos\frac{1}{x}}{\cos x}}$$
78697
I came accross: $$\lim_{x \to\ 0}\frac{x\cos x - \log (1 + x)}{x^{2}}$$  I tried the following please tell me where I 'm wrong:  $$\lim_{x \to\ 0}\frac{x\cos x - \log (1 + x)}{x^{2}} $$  $$\text{(Dividing by }x)$$   $$=\displaystyle\lim_{x \to\ 0}\dfrac{ \cos x - \dfrac{\log (1 + x)}{x}}{x} $$  $$=\lim_{x \to\ 0}\frac{\cos x - 1}{x} $$  $$=\lim_{x \to\ 0}\frac{-2\sin^{2} \dfrac{x}{2}}{x} $$  $$=\lim_{x \to\ 0}\frac{-2x\sin^{2}\dfrac{x}{2}}{(\dfrac{x}{2})^{2}\times 4} $$ $$\lim_{x \to\ 0}\dfrac{-x}{2}=0$$ But Answer given is $\dfrac{1}{2}$  Please Help.
101377
I am trying to find the limit of   $$\lim_{x\to 0}\frac{\sin{3x}}{x}$$  I have no idea what I am supposed to do. I know the identity that,  $$\lim_{x\to0}\frac{\sin{x}}{x} = 1$$  but that will not be good enough on a test and I am not sure why that is true anyways. I do not know how I am supposed to proceed with this problem.
13175
I have $$\lim \limits_{x\to 0}  \frac {\tan(2x)}{\sin(x)}$$ and in my case the result is $\frac{2}{1}$ =2 not whether it is right.  This is my procedure.  $$\lim \limits_{x\to 0}  \frac{\frac {\sin(2x)}{\cos(2x)}}{\frac{\sin(x)}{1}}= \lim \limits_{x\to 0}  {\dfrac {\sin(2x)}{(\cos(2x))(\sin(x))}}=\dfrac{2x\frac {\sin(2x)}{2x}}{\cos(2x)\frac{x\sin(x)}{x}}$$  I separate the limit.  $$\frac{\left(\lim \limits_{x\to 0}2x\right) \cdot \left(\lim \limits_{x\to 0}\frac {\sin(2x)}{2x}\right)}{\lim \limits_{x\to 0}\left(\cos(2x)\right)\cdot\left(\lim \limits_{x\to 0}\frac{x\sin(x)}{x}\right)} = \lim \limits_{x\to 0} \dfrac{2x}{x}=\frac{2}{1} =2$$
91542
Determine the limit:  $$\lim_{h\to0} \frac{\cos(x+h) - \cos(x)}{(x+h)^{1/2} - x^{1/2}}$$  After taking the conjugate, I got:  $$\lim_{h\to 0} \frac{\big(\cos(x+h) - \cos(x)\big)\big((x+h)^{1/2} + x^{1/2})\big)} h$$  I took the conjugate of this, but I don't see how I can cancel out the $h$.   Any tips?