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Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.
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Chapter 4 Solutions
Bundle: Single Variable Calculus: Early Transcendentals, 8th + WebAssign Printed Access Card for Stewart's Calculus: Early Transcendentals, 8th Edition, Multi-Term
- Use the Bessel Function equation to show that: 1₁(x) X lim x0 = (3.1)arrow_forward2. Consider the function g(x) = x cot (3x). 3x A. Function g can be rewritten as · cos(3x) · ( 5). Explain. sin (3x) B. Rewriting function g as in Part A, it follows that limx→0 g(x) = . Explain.arrow_forwardIf 2 - x s g(x) s 2 cos x for all x, find lim,0 8(x).arrow_forward
- Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. sin(x – 1) lim x-1 x3 + 4x – 5arrow_forwardA graphing calculator is recommended. Use the Squeeze Theorem to show that lim x→0 x2 cos(16?x) = 0. Illustrate by graphing the functions f(x) = −x2, g(x) = x2 cos(16?x), and h(x) = x2 on the same screen. Let f(x) = −x2, g(x) = x2 cos(16?x), and h(x) = x2. Then ≤ cos(16?x) ≤ ⇒ ≤ x2 cos(16?x) ≤ . Since lim x→0 f(x) = lim x→0 h(x) = , by the Squeeze Theorem we have lim x→0 g(x) = .arrow_forwardEvaluate each limit or explain why it does not exist. Assign OO or appropriate. You are not allowed to use L'Hospital's Rule for any part of this question. ∞ when x² – 9 a. lim I→3 2x2 tan(3x) sin(2x 7x + 3 b. lim x² 3x? 12 c. lim I→-2 x + /6+ xarrow_forward
- Find the limits sin (x? – 4) lim x - 2arrow_forwardLet f(x) = -8x + 3x². Then the expression can be written in the form Ah + Bx+C, where A, B, and C are constants. (Note: It's possible for one or more of these constants to be 0.) Find the constants. А B - = C= = Use your answer from above to find lim h→0 lim h→0 f(x + h) - f(x) h = Finally, find each of the following: ƒ'(1) : ƒ'(2) = ƒ'(3) = f(x+h)-f(x) h f(x+h)-f(x) harrow_forward8 Explain why the form 1 is indeterminate and cannot be evaluated by substitution. Explain how the competing functions behave. Choose the correct answer. OA. If lim f(x)=1 and lim g(x)=co, then f(x)9(x)1 as xa, which is meaningless. Therefore, substitution cannot be used. xa x-a OB. If lim f(x)=1 and lim g(x)=co, then f(x)9(x) as xa, so the limit does not exist. Therefore, substitution cannot be used. x-a xa OC. If lim f(x)=1 and lim g(x) = oo, then f(x)oo as x-a, so the limit does not exist. Therefore, substitution cannot be used. x-a x-a D. If lim f(x)=1 and lim g(x) co, then f(x)9(x)0° as x-a which is meaningless. Therefore, substitution cannot be used. x-a х-аarrow_forward
- 3. Find the value of : sin3x lim x→0 2xarrow_forwardPlot the function and use the graph to estimate the value. lim 2x − cos x/xx→0arrow_forwardA graphing calculator is recommended. Use the Squeeze Theorem to show that lim (x cos(23rx)) = 0. Illustrate by graphing the functions f(x) = -x, g(x) = x2 cos(23Tx), and h(x) = x2 on the same screen. Let f(x) = -x2, g(x) = x2 cos(23nx), and h(x) = x2. Then ? vs cos(23nx) S ? v = ? vsx cos(23TX) < ? V Since lim f(x) = lim h(x) = by the Squeeze Theorem we have lim g(x) = x-0 x-0 x-0arrow_forward
- College AlgebraAlgebraISBN:9781305115545Author:James Stewart, Lothar Redlin, Saleem WatsonPublisher:Cengage LearningCalculus For The Life SciencesCalculusISBN:9780321964038Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.Publisher:Pearson Addison Wesley,
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