Exercise 13. Show the following results.1+tan x1+tanh1/sina(a) limx-0(b) lim (2+3 - 12) tan(x/4 = (24.39)-4/= 1(c) lim (sin x)¹/(2x-π = 1x→n/2(d) lim2+1+ ln(1x-1√x² - 1)= 0

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Answered: Exercise 13. Show the following…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Q16 Needed to be solved this multiple choice question correctly in 20 minutes in the order to get positive feedback please show me neat and clean work for it by hand solution needed needed Hundred percent efficiency needed
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b) lim 2x² + cos(3x) x’−10x+cosx
9. lim(4sec x-4 tan x) | X-+
Q1) Show that: (a) (1+x)° =1+6x (b) :(cosx): = -sinx dx (e) (V) = d 1 1 ( VI) = (d) /1+ tan x = 1+: dx 2 Vx 2x2+2x+1 x² –1 Q2) Find the asymptotes for the following functions : f(x)=- g(x) = 1+x2 2x V1+cos 2x (1) lim,- 4(x- |4-x2| Q3) Find: (2) limx→-2- (x+ 2) 2-x Q4) Find the equation of the normal to the curve [y+x(1+y)-4] at point (2,1).
(f) lim x-0 (sin x)(arctanx) -x 1 - cos(x2)
a2
16. lim cos²x-1 x-0 1-cos x
13) The integer n = a) If n = 3 (mod 9), then what is x ? b) If n = 3 (mod 11), then what is x ? Hint for #13: 12700140x3243243 is missing the digit x. Consider the number 1234. By writing each digit in scientific notation, we have 1234 1 x 10³ +2× 10² +3 × 10¹ + 4 × 10⁰ Then notice that 10 = 1 (mod 9). When we reduce mod 9, we obtain 1234 1 x 10³ +2× 10² +3 x 10¹ + 4 x 10⁰ = = 1x 1³ + 2 x 1² + 3 × 1¹ + 4 × 1⁰ = 1+ 2+ 3+ 4 = 10 = 1 (mod 9). Since 10 = -1 (mod 11), when we reduce mod 11, we obtain 1234 1 x 10³ +2× 10² +3 x 10¹ + 4 x 10⁰ = = 1x (-1)³ +2× (-1)² +3 × (-1)¹ +4× (-1)⁰ = -1 + 2-3 + 4 = 2 (mod 11).
Evaluate: lim X-OL sin(6x) sin(4x) 2 A 3 (в) о C) 1 3 D 2
Q2) Find : (a) lim (2 + x) 16-x²| 1-secx cos2x sintx (b) lim (c) lim 0 sinx (d) lim 4-x cosx-sinx X0 sin2x cos2x Q3) Graph: fi(x)=2+sin2x, f(x)=14-x , f(x)=1-2cos , fx)-V4- x, Ss(x)= 1+x

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Exercise 13. Show the following results. 1/sin z 1 + tan x (a) lim I-0 1+tanhr = 1 (b) | lim(2* +3* – 12)tan(=z/4 = (2*.39)—4/* (c) lim (sin x)¹/(2x-* = 1 I-R/2 2² - 1 z+¹+ In(1-√√x² - 1) (d) lim = 0

Exercise 13. Show the following results. 1/sin z 1 + tan x (a) lim I-0 1+tanhr = 1 (b) | lim(2* +3* – 12)tan(=z/4 = (2.39)—4/ (c) lim (sin x)¹/(2x-* = 1 I-R/2 2² - 1 z+¹+ In(1-√√x² - 1) (d) lim = 0