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3. lim x 3 (x-3) tan(9-x²) sin (2x-6)

3. lim x 3 (x-3) tan(9-x²) sin (2x-6)

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter4: Calculating The Derivative
Section4.2: Derivatives Of Products And Quotients
Problem 37E
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[question#1_partB] Evaluate the following limits. (Example solution is provided in the photo, kindly follow the format). USE OF L’HOPITAL’S RULE IS NOT ALLOWED.

 

Evaluate the following limits. USE OF L'HOPITAL'S RULE IS NOT ALLOWED. 3. lim (x-3) tan(9-x²) sin² (2x-6) x 3 log_₁(x+3) 1-cosh (cos¹ x) 4. lim x →1
Transcribed Image Text:Evaluate the following limits. USE OF L'HOPITAL'S RULE IS NOT ALLOWED. 3. lim (x-3) tan(9-x²) sin² (2x-6) x 3 log_₁(x+3) 1-cosh (cos¹ x) 4. lim x →1
fa) - 2u sinh (x-1) ta - X = Note: DOM F(x) E IR f'(x) = (2xanh (x-1) +2²-x) f'(x) = d (2xsinh (x-1) + = (2x) - 2 / (x) dy dx f'(x) = 2x coch (x-1) + 2² In (2) + 24h(x-1)- / ('(x) is continuous on (0,1), therefore the function is differentiable on (0,1) f(0) = 2 (0) 6inh (0-1) +2°-0 f(₁) = 2(1) sinh (1-1) +2'-1 Fli) - 2 sinh lo) t flo) = 0+1 -0 flo) = 1 there is no x value parallel to the line a=0 b=1 7 where the tangent line at xis that passes through the end points lim x + -2²¯ f(x)= x³-4x² (x+2) ² (-2) ³ -4 (-2) ² (-2+2) ² 8-4 (9) O = RHL since to re LAL CLONT asyMPTOTES: lim x +∞ x² +4x +4 llence y = x- 8 there fore there is a vertical J at x = -2 (f(x)), lim x++oo (f(x)-ax) x³-4y² +x X³-√x²-x - 8x² -Ux lim x + 2+ f(x)= x³-Yv² (x+2) f(x) = (-2)² -4 (-2)² (-2+2)2 O 8y² +324 32 2 8x +32 asymptote
Transcribed Image Text:fa) - 2u sinh (x-1) ta - X = Note: DOM F(x) E IR f'(x) = (2xanh (x-1) +2²-x) f'(x) = d (2xsinh (x-1) + = (2x) - 2 / (x) dy dx f'(x) = 2x coch (x-1) + 2² In (2) + 24h(x-1)- / ('(x) is continuous on (0,1), therefore the function is differentiable on (0,1) f(0) = 2 (0) 6inh (0-1) +2°-0 f(₁) = 2(1) sinh (1-1) +2'-1 Fli) - 2 sinh lo) t flo) = 0+1 -0 flo) = 1 there is no x value parallel to the line a=0 b=1 7 where the tangent line at xis that passes through the end points lim x + -2²¯ f(x)= x³-4x² (x+2) ² (-2) ³ -4 (-2) ² (-2+2) ² 8-4 (9) O = RHL since to re LAL CLONT asyMPTOTES: lim x +∞ x² +4x +4 llence y = x- 8 there fore there is a vertical J at x = -2 (f(x)), lim x++oo (f(x)-ax) x³-4y² +x X³-√x²-x - 8x² -Ux lim x + 2+ f(x)= x³-Yv² (x+2) f(x) = (-2)² -4 (-2)² (-2+2)2 O 8y² +324 32 2 8x +32 asymptote
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