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Limits Evaluate the following limits.
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Calculus: Early Transcendentals (3rd Edition)
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- ax? + sin? y 96x2 + y² Let a be a nonzero constant such that a + 96. Let f(x, y) and let L= lim(r.y)→(0,0) F(x, y). Which of the following statements is correct? ax² + sin y Türkçe: a sıfırdan farklı ve a + 96 olsun. Eğer f(x, y) ve 96x2 + y² L= lim(r,y)→(0,0) F(x, y) ise aşağıdaki cümlelerden hangisi doğrudur? O L does not exist because the limit as (x,y) goes to (0,0) along the x-axis is different from the one along the y-axis. (x-ekseni ve y-ekseni üzerinde limitler farklı olduğu için L yoktur.) O L=1 O L=a/96 O L does not exist because f is undefined at (0,0). (f fonksiyonu (0,0) noktasında tanımsız olduğu için L yoktur). O L exists if a=97 because in this case the limit as (x,y) goes to (0,0) along x=0 is the same as the one along y=x. (Eğer a=97 ise limit vardır çünkü bu durumda x=0 ve y=x doğruları üzerinde limitler aynıdır.)arrow_forwardHow do I find this limit?arrow_forwardt2 Find the limit: lim (e-3t i+ j+ cos 2t sin?tarrow_forward
- ax² + sin? y Let a be a nonzero constant such that a + 18. Let f(x, y) and let 18x? + y² lim(z,y)→(0,0) F(x,y). Which of the following statements is correct? ax? + sin? y Türkçe: a sıfırdan farklı ve a 18 olsun. Eğer f(x, y) ve 18x? + y? L = lim(m,y)→(0,0) f(x, y) ise aşağıdaki cümlelerden hangisi doğrudur? O L does not exist because the limit as (x,y) goes to (0,0) along the x-axis is different from the one along the y-axis. (x-ekseni ve y-ekseni üzerinde limitler farklı olduğu için L yoktur.) O Lexists if a=19 because in this case the limit as (x,y) goes to (0,0) along x=0 is the same as the one along y=x. (Eğer a=19 ise limit vardır çünkü bu durumda x=0 ve y=x doğruları üzerinde limitler aynıdır.) O L=1 O L does not exist because f is undefined at (0,0). (f fonksiyonu (0,0) noktasında tanımsız olduğu için L yoktur). O L=a/18arrow_forwardax? + sin? y Let a be a nonzero constant such that a + 29. Let f(x, y) and let 29x2 + y? L= lim(",y)→(0,0) F(x, y). Which of the following statements is correct? ax² + sin? y Türkçe: a sıfırdan farklı ve a 29 olsun. Eğer f(x, y) ve 29x2 + y² L = lim(z,9)>(0,0) f(x, y) ise aşağıdaki cümlelerden hangisi doğrudur? O L=a/29 O L exists if a=30 because in this case the limit as (x,y) goes to (0,0) along x=0 is the same as the one along y=x. (Eğer a=30 ise limit vardır çünkü bu durumda x=0 ve y=x doğruları üzerinde limitler aynıdır.) O L does not exist because the limit as (x,y) goes to (0,0) along the x-axis is different from the one along the y-axis. (x-ekseni ve y-ekseni üzerinde limitler farklı olduğu için L yoktur.) O L does not exist because f is undefined at (0,0). (f fonksiyonu (0,0) noktasında tanımsız olduğu için L yoktur). O L=1arrow_forwardHW6 Q3arrow_forward
- x²– y2 )2 Consider the function f (x,y) = Which ONE of the following statements x2+ y2 is TRUE? O A. f (x, y) exists and it is equal to 1 as verified along x– axis and y – axis; lim (л.у) — (0,0) B. The function f is not continuous at point (0,0); OC. lim f (x, y) exists and it is equal to 0; (л.у) — (0,0) D. The function f is continuous in the whole domain; O E. None of the choices in the list.arrow_forwardUse the graphing utility to plot the curve y = sin (x) and the horizontal lines y = 1 and y ly%3D (When using the graphing utility, separate each expression with a comma.) y = 合 1- 2 4 6. -1-arrow_forwardEvaluate using the Squeeze Theorem. lim cos θ cos(tan θ) θ→ π/2arrow_forward
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning
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