here they are

in Calculus Answers by Level 2 User (1.4k points)
reopened by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

Best answer

(1) Let x=4+h, then we have (3-√(4+h+5))/h, (3-√(9+h))/h=(3-√(9(1+h/9)))/h,

(3-3√(1+h/9))/h=(3-3(1+h/9)^½)/h=(3-3(1+h/18))/h when h is small.

This becomes (3-3-h/6)/h=-⅙ as h→0, that is, in the limit as x→4.

(2) Factorise:

(x-3)(x+3)(x²+9)/((2x+1)(x-3))=(x+3)(x²+9)/(2x+1) when x≠3.

Therefore limit as x→3 is (3+3)(9+9)/(6+1)=6×18/7=108/7.

(3) √(3x-5)/5=√(30-5)/5=√25/5=5/5=1 as x→10, by directly putting x=10 in the expression, because there are potential anomalies here.

(4) Let x=1+h, then 

((1+h)^⅓-1)/((1+h)^¼-1)=(1+h/3-1)/(1+h/4-1)=(h/3)/(h/4)=4/3 as h→0, x→1.

(5) (cos(2x)-1)/(cos(x)-1)=(2cos²(x)-1-1)/(cos(x)-1)=


2(cos(x)+1) as x→0, which is 2(2)=4.

(6) No potential anomaly here, so just put x=3:



by Top Rated User (1.0m points)
selected by

Related questions

1 answer
asked Apr 12, 2020 in Calculus Answers by qwertykl Level 2 User (1.4k points) | 1.1k views
1 answer
asked Apr 8, 2020 in Calculus Answers by qwertykl Level 2 User (1.4k points) | 512 views
1 answer
1 answer
asked Apr 5, 2014 in Calculus Answers by amandawho Level 1 User (620 points) | 343 views
1 answer
0 answers
1 answer
asked Aug 15 in Calculus Answers by Audrey Matey Level 1 User (140 points) | 37 views
Welcome to, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
87,067 questions
96,685 answers
24,366 users