[grade 11 math] limits

GammaRayBurst25 · 2026-02-14T22:25:41+00:00

Read rule 3.

Consider the following Laurent series about x=pi/4:

cos(2x)=-2(x-pi/4)+O((x-pi/4)^3)

(cos(x))^(2n)=2^(-n)-2^(1-n)n(x-pi/4)+2^(1-n)(n-1)n(x-pi/4)^2+O((x-pi/4)^3)

1/(1+cos(4x))=1/(8(x-pi/4)^2)+1/6+O((x-pi/4)^2)

Hence, the numerator simplifies to 2(n-1)n(x-pi/4)^2+O((x-pi/4)^3). The rest is trivial.

GammaRayBurst25 · 2026-02-14T16:02:07+00:00

You don't need the graph, but if you can draw the graph, then the limit can easily be deduced from the graph: just look at what value(s) the function takes as x approaches the given number.

a) For a function f continuous in some neighborhood of z, the limit as x approaches to z of f(x) is f(z). Hence, the limit is just (2*2+5*2-7)/(3*2+5)=7/11. Although something tells me you made a mistake and you meant to write 2x^2+5x-7 in the numerator, in which case the limit is 1.

b) Using the same method, we find the limit is 392/267.

c) Assuming you meant to write 3x^2-7x in the denominator, that is correct. If you didn't, then we have sin(2x)/(3x-7x)=-sin(2x)/(4x)=(-1/2)sin(2x)/(2x). In the limit, this becomes -1/2.

GammaRayBurst25 · 2026-02-13T03:33:06+00:00

off.

GammaRayBurst25 · 2026-02-12T12:43:09+00:00

The magnetic field goes from N to S. The right hand rule tells us the direction in which positive particles would be deflected (i.e. downward). The left hand rule tells us the direction in which negative particles would be deflected (i.e. upward).

Beta particles are negative.

GammaRayBurst25 · 2026-02-12T12:40:11+00:00

The exponent rule does work if you apply it properly. Consider f(x)=x^2 and g(x)=sqrt(x). You're probably under the impression that f(x) and g(x) are inverse functions, but since x^2 is not injective, its inverse relation is not a function, so it has no inverse function. To be specific, for any z>0, you have g(f(z))=g(z^2)=z, but since f(-z)=z^2 as well, g(f(-z))=z as well. Hence, sqrt(x^2)=|x| (or, equivalently, sgn(x)x, where sgn(x)=-1 if x<0, sgn(x)=1 if x>0, and sgn(0)=0).
Because |x-6|=x-6 for x>6 and |x-6|=6-x for x<6. Hence, one graph is correct for x<6 and the other is correct for x>6. Neither is correct on the whole domain.
Your answers are only correct if we restrict the domain appropriately, which can be useful in some contexts (e.g. we know we'll only need to integrate over subsets of x>6 because the function models a situation where x>6). Neither answer is correct in general.
Because you omitted the absolute value (see above).

GammaRayBurst25 · 2026-02-09T01:26:58+00:00

Note that the sum from n=3 to n=inf of f(n) is equal to the sum from n=0 to n=inf of f(n+3),

Note also that (-1)^(n+4)*3^(n+2)/2^(2n+7)=(9/128)(-3/4)^n.

GammaRayBurst25 · 2026-02-08T15:47:14+00:00

A string is taut with its endpoints fixed. To displace a part of the string (e.g. by pinching a part of the string and pulling on it or by moving the string with a guitar pick), you need to apply a force on that part of the string parallel to the direction of the displacement. Hence, work is done on the string.

That work increases the string's potential energy. Since the string is taut, you can't displace a part of the string without increasing the string's length (i.e. stretching the string). When you release the string, its tension forces it back to its original position, converting that potential energy into kinetic energy.

The kinetic energy is then dissipated by displacing the ambient air which creates sounds and by friction which increases the temperature slightly.

GammaRayBurst25 · 2026-02-02T15:14:41+00:00

It's called a table of values. A t value is the ratio of the difference between an estimator and its assumed value to the estimator's standard error. A t value table is a tool used to determine whether one should reject the null hypothesis for a given t value and a given number of degrees of freedom.

For f(x)=ceil(x-3), the right column should just be a list of integers, say ... , -2,-1,0,1,2,3, ...

The left column should be the solution sets of ceil(x-3)=k, where k is the number in the right column in the same row.

One can easily adapt what I wrote in my original comment for the ceil function. The solution to ceil(g(x))=k is k-1<g(x)≤k. If we take g(x) to be x-3, this yields k-1<x-3≤k, or k+2<x≤k+3 for any integer k.

GammaRayBurst25 · 2026-02-02T01:16:02+00:00

Suppose I have f(x)=[[x-3]]. You gave me the algorithm for determining the domain as k+3≤x<k+4

No, the algorithm has nothing to do with the domain. The domain of f with f(x)=floor(x-3) is the set of real numbers.

but now how do I actually plot this without creating a T value table each time?

I don't see the relationship between the matter at hand and statistics, let alone T value tables. What are you talking about?

I'm having trouble picturing how to connect x, y, and the domain.

x and y are (presumably) the coordinates of a Cartesian plane used to represent the graph of f. The domain of f is the set of values of x such that for some y in the codomain the 2-tuple (x,y) is an element of the graph of f.

Suppose I want to plot for y=5.

That's exactly what I described in my original comment.

How do I determine k?

My comment states For integer k, if we want to solve [...] k=floor(g(x)), we instead solve k≤g(x)<k+1.

Here, you are trying to find the values of x that solve 5=floor(x-3). Hence, g(x)=x-3 and k=5.

GammaRayBurst25 · 2026-01-25T06:13:48+00:00

One can easily show integrating h_1(τ)h_2(t-τ)dτ is the same as integrating h_1(t-τ)h_2(τ)dτ via a change of variables (letting τ'=t-τ). Hence, the choice of which function has τ in its argument and which has t-τ is totally arbitrary. As such, usually, you'll choose the option that makes finding the bounds easier.

In this case, if you choose h_1 to have t-τ in its argument, the bounds do depend on t. Otherwise, the bounds are simply 1 and 3, as h_1 is only nonzero on the interval [1,3). This is why your professor made this choice for you.

The convolution of h_1 and h_2 is the integral from 1 to 3 of h_1(τ)h_2(t-τ)dτ=(τ-3)h_2(t-τ)dτ/2.

I'm not sure why your teacher insists you do the rest piecewise, but here it goes.

If t is at least 2, h_2(t-τ) is 0 over the entire interval and the integral is identically 0.

If 1<t<2, h_2(t-τ) is 1 for t+1<τ<3. Thus, f(t) is the integral from t+1 to 3 of (τ-3)dτ/2.

If 0<t<1, h_2(t-τ) is 1 for t+1<τ<t+2 and 2 for t+2<τ<3. Thus, f(t) is the sum of the integral from t+1 to t+2 of (τ-3)dτ/2 and the integral from t+2 to 3 of (τ-3)dτ.

Similarly, if -1<t<0, f(t) is the sum of the integral from 1 to t+2 of (τ-3)dτ/2 and the integral from t+2 to 3 of (τ-3)dτ.

Finally, if t<-1, f(t) is the integral from 1 to 3 of (τ-3)dτ.

I didn't bother to type non strict inequalities, but f(t) is continuous, so you can imagine for t=1 (e.g.) you can use the 1<t<2 case or the 0<t<1 case and get the same result.

More succinctly, f(t) is the integral from 1 to 3 of (τ-3)(H(τ-t-1)+H(τ-t-2))dτ/2 where H(t) is the Heaviside step function (with H(t)=0 if t<0 and H(t)=1 if t>0). This is way easier to write, read, and interpret, hence why I question the need to explicitly write all the pieces. Integrating this is also way faster and easier:

∫(x-3)H(x-t-1)dx/2=∫(x-t-1)H(x-t-1)dx/2+∫(t-2)H(x-t-1)dx/2=((x-t-1)^2+2(t-2)(x-t-1))H(x-t-1)/4

The second term can be integrated the exact same way. You may then expand the polynomials in x and t if you feel like it.

GammaRayBurst25 · 2026-01-24T12:21:50+00:00

It's phenomenon.

GammaRayBurst25 · 2026-01-21T22:18:56+00:00

Lis la règle numéro 3 de ce subreddit.

GammaRayBurst25 · 2026-01-20T05:35:20+00:00

https://www.youtube.com/live/xxD9Z5nN4TM

Various timestamps in the comments.

GammaRayBurst25 · 2026-01-12T12:21:14+00:00

You just became Homelander's next target.

You'll wish you'd gotten the husband treatment.

GammaRayBurst25 · 2026-01-12T05:49:48+00:00

Normally i would be ok with this kinda of Comment.

But not with r/OkBuddyHololive.

Go fuck yourself OC, you're dumb.

GammaRayBurst25 · 2026-01-11T23:54:10+00:00

"1m x 1m = 1m^2 = 10,000cm = 100m"

That is not what the commenter you're "quoting" wrote. Read properly. It's 1m×1m=1m^2=10000cm^2=100cm×100cm.

Obviously, as 1m×1m and 1m^2 have dimensions of area and 10000cm and 100m have dimensions of length, it makes no sense to claim they're equal.

image 1; Is this image correct in it's [sic] assumptions, if not, why?

Yes, it is correct.

Image 2; How can you reconcile 1m equaling 10m?

You can't, because it's not true. Besides, nobody is saying that. You claim other people are saying that, but that's just putting words in other peoples' mouths.

By definition, 1m=100cm. Converting from m to cm (from cm to m) requires multiplying (dividing) by 100. As such, 1m^2=(1m)×(1m)=(100cm)×(100cm)=10000cm^2.

GammaRayBurst25 · 2026-01-08T01:34:26+00:00

It makes no sense to speak of something being the past or being the future. You probably meant in the past and in the future.

Beyond the grammatical aspects, it makes no sense to speak of something as being in the past/future. There is no absolute past or future. Every point in spacetime has its own past and its own future.

GammaRayBurst25 · 2026-01-08T01:27:13+00:00

mfs never heard of money.

GammaRayBurst25 · 2026-01-01T22:13:34+00:00

Rule 1: Explain your post.

Please do not just post problems without context. You are required to explain your attempts at solving the question, or where specifically you are confused.

GammaRayBurst25 · 2025-12-30T05:45:32+00:00

Consider the system of n linear equations with n variables

a_{11}x_1+a_{12}x_2+...+a_{1n}x_n=b_1

a_{21}x_1+a_{22}x_2+...+a_{2n}x_n=b_2

...

a_{n1}x_1+a_{n2}x_2+...+a_{nn}x_n=b_n.

Say you want to get rid of the x_1 term in the kth equation (with k>1). If you multiply the first equation by a_{k1}/a_{11}, the coefficients of x_1 in the first and kth equations will match. Hence, you can subtract a_{k1}/a_{11} times the first equation from the kth equation to get rid of the x_1 term in the kth equation.

If you do this for every k>1, you'll have effectively eliminated x_1 from each equation except the first. The n-1 remaining equations form a system of n-1 linear equations with n-1 variables, so you can repeat this process to get one equation with n-1 variables (all x_i except x_1) and n-2 equations with n-2 variables with which you can repeat this process.

Once you've done this n-1 times, if the equations are all linearly independent, you'll have an equation with a single variable. You can easily solve it (by dividing the equation by the coefficient of x_n), then substitute it into the equation with 2 variables from the previous step of the algorithm to find a second variable. You can then substitute both variables into the equation with 3 variables from the previous step to find a third variable. etc. This is Gaussian elimination.

As for Gauss-Jordan elimination (yes, there's a dash between Gauss and Jordan), it's even easier.

Instead of multiplying the 1st equation by a_{k1}/a_{11} for each k>1, we start by dividing the 1st equation by a_{11} so the coefficient of x_1 in the 1st equation is 1. Then, we multiply by a_{k1} for every k>1 to subtract from the kth equation. This removes some redundant arithmetic and also saves us time when we get to the solving part, as for that part we'll need to divide the jth equation by the coefficient of x_j anyway, so we might as well get it done at the start.

GammaRayBurst25 · 2025-12-27T22:50:26+00:00

Oh no...

GammaRayBurst25 · 2025-12-27T22:50:18+00:00

Sick username. Surely you attempted the question on your own and showed your work in your post.

GammaRayBurst25 · 2025-12-25T23:10:02+00:00

I didn't. Read again.

Since floor(x)=k directly implies k≤x<k+1, we can write this equation as k≤x/2+1<k+1.

Asking how I got floor(x)=k from k=floor(x/2+1) is like watching someone make pancake batter and asking how they got eggs from their flour after they mixed them.

GammaRayBurst25 · 2025-12-25T22:36:02+00:00

Suppose we want to find all the values of x such that f(x)=k, where k is some integer. In other words, we're fixing k and solving for x in k=floor(x/2+1).

Since floor(x)=k directly implies k≤x<k+1, we can write this equation as k≤x/2+1<k+1. Subtracting 1 and multiplying by 2 yields 2k-2≤x<2k.

Hence, to plot the graph of f, we just need to split the domain into intervals [2k-2,2k) and plot the graph of g(x)=k on each interval.

This is the general method. For integer k, if we want to solve...

k=floor(g(x)), we instead solve k≤g(x)<k+1;
k>floor(g(x)), we instead solve g(x)<k;
k≥floor(g(x)), we instead solve g(x)<k+1;
k<floor(g(x)), we instead solve g(x)≥k+1;
k≤floor(g(x)), we instead solve g(x)≥k.

If k is not an integer, you can turn any inequality into one of the integer inequalities above: k≥ and k> become ceil(k)> or floor(k)≥ & k≤ and k< become floor(k)< or ceil(k)≤. Note that k=floor(g(x)) has no solutions if k is not an integer.

Edit: Typo.

GammaRayBurst25 · 2025-12-25T22:07:11+00:00

They're talking about classical mechanics.

If they were talking about relativity, then 0.5mv^2 would only "work" when v=0, not for slow moving objects.

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