Few limit questions?
1) derive f(x)=25/(2x+3) f'(x)=-25(2)/(2x+3)^2 f'(x)= -50/(2x+3)^2 now find f'(1) f'(1)= -50/(2+3)^2 = -50/25 = -2 2) if you substitute t by 2 you end up with 0/0 (indefinite), use L'Hopital theory (derive the top and derive the bottom, then replace t with 2) lim (4t-3)/(2t+1)=5/5= 1 t-->2 3) lim x^2/|x|=lim x^2/x=lim x= 0 x-->0+ x-->0+ x-->0+ limx^2/|x|=lim x^2/ (-x)=lim (-x)=0 x-->0- x-->0- x-->0- So the limit exists and are equal---> zero 4) lim 1/|x|=1/x= + infiniti ---> limit does not exist x-->0+ lim x-->0+
Help with a calculus limit question?
ok, multiply everything out. so we have. h(x) = (kxx - 2kx - 3x + 6) / (14xx + 35x + 18x + 45) now, when we're talking about horizontal asymptotes, we're worried about the leading coefficients of the term with the highest degree. in this case, the term with the highest degree in both the numerator and deonominator is xx or x^2. the leading coefficient on top is k and the leading coefficient on the bottom is 14. so, in order to have a horizontal asymptote at y = 2 k / 14 = 2 cause as x goes to infinity, all the other terms don't really matter and only xx will dominate. so, the leading coefficients of the xx will determine the horizontal asymptote. multiply both sides by 14 and you get k = 28. graph on you're calculator just to make sure. =)
Help needed on Limit Questions!!!?
1) f(x) = [(x^2+9x+18)^0.5]/x^2-9 f is continuous at x=0, and f(0) = - sqrt2 / 3 so f(x) -> f(0) when x->0 2) [x*Sin(8x)]/[1-Cos(8x)] Note that cos8x = (cos4x)^2 - (sin4x)^2 = 1 - 2(sin4x)^2, so 1 - cos8x = 2(sin4x)^2 Also, sin8x = 2(sin4x)(cos4x). Therefore: [x*Sin(8x)]/[1-Cos(8x)] = x sin4x cos4x / (sin4x)^2 = (1/4) 4x cos4x/sin4x = (1/4) cos4x / [(sin4x)/(4x)] But we know that (sin4x)/(4x) -> 1 as x->0 (well-known result), and cos4x -> 1 as x->0, so [x*Sin(8x)]/[1-Cos(8x)] -> 1/4 as x->0 3) lets mutiply [sqrt(1+x) - 1]/Sin(x/2) by sqrt(1+x) + 1, and divide by the same expression: [sqrt(1+x) - 1]/Sin(x/2) = [sqrt(1+x)-1]/Sin(x/2) * [sqrt(1+x)+1]/[sqrt(1+x)+1] = [(1+x)-1] / [sin(x/2) (sqrt(1+x)+1)] = x / [sin(x/2) (sqrt(1+x)+1)] = { 2 / [sin(x/2) / (x/2)] } * { 1/ (sqrt(1+x)+1) } Now [sin(x/2) / (x/2)] -> 1 as x->0, and sqrt(1+x)+1 -> 2 as x->0, so [sqrt(1+x) - 1]/Sin(x/2) -> (2/1) * 1/2 = 1 as x->0 4) i'm unsure if you made a typo: is it really (sinx^3), or did you want to write (sinx)^3 ? In any case, the limit doesn't exist, you'd still have 1/x^2 that would explode to infinity as x->0.
Stuck on two limit questions?
lim x→0 [1-cos(7x)] / [ x sin (x)] first I suggest to moltiply both numerator and denominator by x²: lim x→0 [(1-cos7x) / x²] [ x² /(x sinx)] = lim x→0 [(1-cos7x) / x²] (x / sinx) now I suggest to moltiply and dividing by 49 so that limit becomes: lim x→0 49 [(1-cos7x) / 49 x²] (x / sinx) now you should remember the known limits: lim x→0 (1- cos x) / x² = 1/2 and lim x→0 (sin x) / x = 1 and, similarly, lim x→0 [(1-cos7x) / 49x²] = lim x→0 (1 - cos 7x) / (7x)² = 1/2 lim x→0 (x / sinx) = lim x→0 1 / [(sinx) / x] = 1 thus the limit becomes: lim x→0 49 [(1-cos7x) / 49 x²] (x / sinx) = 49 (1/2) (1) = 49 / 2 lim x→0 (sin⁶3x) / (x⁵sin x) in which I'm going to multiply and divide by (3x)⁶= 729x⁶: lim x→0 (sin⁶3x) / (x⁵sin x) = lim x→0 [(sin⁶3x) /(3x)⁶] [(3x)⁶/ (x⁵sin x)] = lim x→0 [(sin 3x) /(3x)]⁶ [(3⁶x) / (sin x)] = lim x→0 [(sin 3x) /(3x)]⁶729 [ x / (sin x)] = lim x→0 [(sin 3x) /(3x)]⁶{729 / [(sin x) /x]} = remembering the previus known limit lim x→0 sinx / x = 1 hence lim x→0 (sin 3x) /(3x) = lim x→0 (sin x) / x = 1 your limit becoming: lim x→0 [(sin 3x) /(3x)]⁶{729 /[(sin x) /x]} = (1⁶)(729 /1) = 729 Bye!
Calculus limit question?
how do i find the limit as x-> 0 of (1 + 12x^2 - 2cosx + cos^2x) / 3x^2? also, limit z -> 0 of (sin 7z)/(sin 3z) i am not sure what trig identities i need to use to factor and reduce these.
Why is there a limit on the number of answers to questions you can submit per day on Quora? I have many more answer requests waiting.
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Calculus Limit Questions HELP NEEDED!!?
HELP IS GREATLY APPRECIATED THANK YOU!! The lim as h->0 tan3(x+h) - tan3x / h a) 0 b) 3sec^2(3x) c) sec^2(3x) 3) 3cot(3x) e) nonexistant If f(x) = e^x, which of the following equals f '(e)?? a) lim as h->0 e^x+h / h b) lim as h->0 e^x+h - e^e / h c) lim as h->0 e^e+h - e / h d) lim as h->0 e^x+h - 1 / h e) lim as h->0 e^e+h - e^e / h lim as h->0 ln(e+h) - 1 / h a) f '(e), where f(x) = ln x b) f '(e), where f(x) = ln x / x c) f '(1), where f(x) = ln x d) f '(1), where f(x) = ln (x+e) e) f '(0), where f(x) = ln x ----------- In addition to the answers, please explain how you solve em, thanks!!
Question about greatest integer function with limits?
Solving this step by step we get lim{x --> 2-} (x - [x]) / [x - 3] = (lim{x --> 2-} (x - [x])) / (lim{x --> 2-} [x - 3]) = N / D This holds since (we assume for now that) both numenator and denominator are defined. N = (lim{x --> 2-} x) - lim{x --> 2-} [x] = 2 - 1 = 1 because [x] = 1 for all 1 < x < 2. Analogously, we find that D = lim{x --> 2-} [x - 3] = -2 because [x - 3] = -2 for all 1 < x < 2. Hence the limit of f(x) as x converges to 2 from the left equals N/D = -1/2.
Why has Quora told me "You have reached your limit for answering questions today"?
It's a rolling 24 hour period. So maybe you wrote a lot late yesterday.Look at your answers on your profile page and you can count how many you've written and approximately when you will be able to submit more.What you can do is write your answer, don't post it, and it will be saved as a draft. Try again 24 hours after your last post.You don't need to try to answer every question.Read existing answers to the question. Exactly what you want to say may have been said already.Or the question may be a common one that will get merged. There may be 25–100+ answers just waiting.Single sentence or single paragraph answers are not really what Quora is looking for. They will just get collapsedDig deep, think a lot, answer well. 10 really good answers are better than 50 quickies.The first couple days you're on Quora they limit you more tightly to give you time to learn about the rules and culture of Quora. Limits on new accounts also prevent people joining just to write spam or troll content.My limit is 50 right now. That seems to be true for those who joined in 2018 or 2019.Some longtime Quorans mentioned their limit is 100.If your answer limit doesn't seem to follow these general guidelines, you can always ask.http://Quora.com/help