hello! can you help me answer numbers 6 and 12, please provide solution. can you also find the limit of limx->0 cos(x+sinx) 3-20 Find the limit.x² - 93. lim e4. lim+ 2x - 3x² - 96.5. lim6. lim+ 2x - 3(4 - 1) + 1hx² + 2x – 331-347. lim8. limState8.31.4 - v9. lim9 (r - 9)*10. lim-4 4 v331Vx + 6x' - 3xVx? - 91. limu' + 5u?12. limis6u33x² – 9. limX* 2x – 614. limX→-* 2x - 61- 2x? – x*- lim In(sin x)16. lim5 + x - 3x*lím (/x² + 4x + 1- x) 18. lim e*-11lim tan '(1/x)20. lim17x² - 3x + 2122 Use graphs to discover the asymptotes of the curve. Thene what you have discovered.3.

3-20 Find the limit. x² - 9 3. lim 4. lim + 2x - 3 x - 9 5. lim + 2x - 3 6. lim x² + 2x - 3 31-3 (- 1)' + 1 7. lim 12-4 State 8. lim 2t-8 31. 4 v 9. lim (r- 9)* 10. lim 4 4 v - 33 Vx + 6-x x'- 3x u - 1 1. lim →1 u' + 5u? 12. lim X-3 is 2. 6u 33 Vx² - 9 . lim X0 2x /x² -9 14. lim X-* 2x - 6 34 1- 2x? – x* - lim In(sin x) 16. lim 5 +x – 3x* lim (Vx² + 4x + 1 – x) 18. lim e*-* 1 lim tan '(1/x). 20. lim x - 3x + 2 +0+ 22 Use graphs to discover the asymptotes of the curve. Then what you have discovered.

3-20 Find the limit. x² - 9 3. lim 4. lim + 2x - 3 x - 9 5. lim + 2x - 3 6. lim x² + 2x - 3 31-3 (- 1)' + 1 7. lim 12-4 State 8. lim 2t-8 31. 4 v 9. lim (r- 9)* 10. lim 4 4 v - 33 Vx + 6-x x'- 3x u - 1 1. lim →1 u' + 5u? 12. lim X-3 is 2. 6u 33 Vx² - 9 . lim X0 2x /x² -9 14. lim X-* 2x - 6 34 1- 2x? – x* - lim In(sin x) 16. lim 5 +x – 3x* lim (Vx² + 4x + 1 – x) 18. lim e- 1 lim tan '(1/x). 20. lim x - 3x + 2 +0+ 22 Use graphs to discover the asymptotes of the curve. Then what you have discovered.

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section4.5: Rational Functions

Problem 8E

See similar textbooks

Similar questions

1. Broadly speaking, what are the Limit Laws, and why do they help us calculate limits?2. In Example 3, there is a step where the function x2 − 1x − 1is changed to x+ 1. But unlike the secondfunction, the first function is undefined at x = 1. How can we justify changing functions this way?3. Explain the flaw in the following calculation:limx→0x2sin(1/x) = (limx→0x2) · (limx→0sin(1/x)) = 0 · (limx→0sin(1/x)) = 0.
Using: limx→4f(x)=7limx→4f(x)=7 and limx→4g(x)=4limx→4g(x)=4, evaluate limx→4f(x)+g(x)3f(x).limx→4f(x)+g(x)3f(x). Limit = Enter DNE if the limit does not exist.
Evaluate the Limits: a) limt->-1(t2+3t+2)/(t2-t-2) b) limx->9 (9x-x2)/(3-x1/2) c) limx->-inf (ex-e-x)/(ex+e-x) Do Not Use L'Hopital's Rule
Evaluate the limit lim n→ ∞ √( 9n^4)+13n^3-5n-10 / -√( 25n^16)+5n^2-9n Also find the asymptotic value of above limit
how did you get y^2-4xy to y-4x when there are three y's on top I input the numbers and got 0 which would make it where the limit does not exist.
limx → 1 (x-2) / (x-1)2 how do I calculate the following limit without using the L'Hospital's rule
find the limit f(x)= (5-x)/x2-25 a) find the x values at which discountous b) which discontinuties are removable
can you explain how you got (x-2)=(t-3) and can you explain how you got the new limits because the question doesnt require all this complicated computation the question to be clear is a) converging to alpha less than 0. b) converging to alpha greater than 0. c) conveging to 0. d)diverging
Evalute the limit of: lim x->0 (x csc 6x)/(cos 14x) The answer must be in simplified form or a fraction
Find the limit of limitx->0= x2+4, x does not equal 0 = -1, x=0
The finite region bounded by x=y^2 and x=6-2y^2 rotated around the line x=8. The region bounded by the x-axis, y=x^2 and y=2-x rotated around the x-axis. The region bounded by the x-axis, y=x^2 and y=2-x rotated around the y-axis.
(Derivatives) 5.12.4) Find Limit ?