Tagged Questions

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A simpler solution to a limit question?

Okay I saw this limit question in Thomas' Calculus 12th Edition: $\lim \limits_{x \to 0} \frac{\tan3x}{\sin8x}$ The answer is $\frac{3}{8}$. I was able to get the correct answer using this ...
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How to prove geometrically the limit $\lim_{x \to 0}\frac{1-\cos{x}}{x}$ using squeeze theorem [duplicate]

Here, I ask how to geometrically prove the limit $$\lim_{x \to 0}\frac{1-\cos{x}}{x}$$ using squeeze theorem. I see so many proofs for the other important limit ...
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A limit with trigonometric functions

$$\lim_{x \to 0} \frac{1-\cos2x}{x\sin(x+4\pi)}$$ .... and then I got ... $$= \lim_{x \to 0} \frac{2\sin^2x}{x \sin x}$$ and I can't insert 0 in that calculation as long as there's x in the ...
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Trigonometric Limit: $\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)$

I cannot figure out how to solve this trigonometric limit: $$\lim_{x\to 0} \left(\frac{1}{x^2}-\frac{1}{\tan^2x} \right)$$ I tried to obtain $\frac{x^2}{\tan^2x}$, $\frac{\cos^2x}{\sin^2x}$ and ...
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How do i solve this limit? It's about $\frac{\sin(x)}{x} = 1$ and $\frac{1-\cos(x)}{x} = 0$ [duplicate]

$$\frac{\sin(x)}{x} = 1$$ and $$\frac{1-\cos(x)}{x} = 0$$ I can't come up with any solution. Could you please help me? Also: $$\lim_{x\to 0} \frac{1-\cos(x)}{\tan(x)\sin(x)}$$
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Finding the limit of $\lim_{v\to180}\frac{360\cos\left(\frac{v}{2}\right)}{180-v}$

I need to find this limit: $\displaystyle\lim_{v\to180}\frac{360\cos\left(\dfrac{v}{2}\right)}{180-v}$, with $v$ in degrees. I have tried to do this: ...
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A limit problem involving repeated cosines

I was playing around on my calculator and I found something interesting:- Let's say I take some value $x$ in degrees and apply the following operation: $cos(cos(cos(cos....(x)))))...)$. This always ...
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Calculate the limit:

I need to calculate: $$\lim_{x\rightarrow \frac{\pi }{4}}\frac{\sin2x-\cos^{2}2x-1}{\cos^{2}2x+2\cos^{2}x-1}$$ I replaced $2\cos^{2}x-1=\cos2x$ and $\cos^{2}2x=1-\sin^{2}2x$, so this limit equals ...
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Calculating limits of trigonometric functions analytically

I am trying to figure out how to solve a certain set of problems using the trigonometric identities for evaluating limits, which are: $\lim\limits_{\theta \to 0} \dfrac{\sin\theta}{\theta}=1$ ...
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$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}$ without using L'Hopital

$$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=?$$ I tried using $\lim_{x\to0} \frac{\sin x}{x}=1$. But it doesn't work :/
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Trigonometric Limits function

I need to find this limit $\lim_{\theta\to \frac \pi2}$ (2 $\theta$ - $\pi$) sec $\theta$ ?
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Evaluate $\lim_{x\rightarrow{\frac\pi2 }} (\sec(x) \tan(x))^{\cos(x)}$ without L'Hôpital's rule

I have tried changing limit to $\lim_{x\rightarrow0}$ and use some trigonometry identity ($\sin^2(x)+\cos^2(x) = 1$ and $\sin (x+\pi/2) = \cos(x)$) but doesn't work I have no idea on how to do this ...
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Calculating the limit of $\sum_{k=1}^{n}\frac{1}{4^k\cos^2\frac{x}{2^k}}$

How i can prove that $$\lim_{n \to\infty}\sum_{k=1}^{n}\frac{1}{4^k\cos^2\frac{x}{2^k}}=\frac{1}{\sin^2(x)}-\frac{1}{x^2}$$.
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Find $\lim_{x\to 0} \frac{\tan16x}{\sin2x}$

Find $\lim_{x\to 0} \frac{\tan16x}{\sin2x}$ I'm a little confused on limit trig. Am i suppose to simplify tan or do I use the derivative quotient rule? Please Help!!!
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Limit as x approaches zero.

$$\lim_{x\to 0}\frac{\sin(2x)\sin(4x)}{x\sin(3x)}=\frac83$$ I know the answer by the means of Wolfram|Alpha, but I don't know the way there. I have yet to learn L'Hôpital's rule so I cannot apply it. ...
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determine the point at which this function is not continuous and state the type of discontinuity.

absolute value of (sin (1/x) determine the point at which this function is not continuous and state the type of discontinuity is it removable, jump, infinite, or none of these?
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Limit of recurring series with arctan

Let $a_{n+1} = \arctan (a_n)$. Find $\lim\limits_{n\to \infty } \, a_n$ if $a_1$ is chosen arbitrarily. I think the answer is 0, but I don't really know how to prove that. Could you give me some ...
Recently I came across a proof using the apparant fact that $$\lim_{x \rightarrow a} \cos^{-1}(x)=\cos^{-1}(\lim_{x \rightarrow a} x)$$ with justification: because arccos(x) is a continuous function. ...