0
How do you solve the limit as x approaches zero of 1 over x minus 4 plus 1 over x plus 4 all over x?
Wiki User
∙ 12y ago
Limx→0 [ 1 / (x - 4) + 1 / (x + 4) ] / x
= Limx→0 1 / (x2 - 4x) + 1 / (x2 + 4x)
= Limx→0 (x2 + 4x) / (x4 - 16x2) + (x2 - 4x) / (x4 - 16x2)
= Limx→0 (x2 + 4x - 4x + x2) / (x4 - 16x2)
= Limx→0 2x2 / (x4 - 16x2)
= Limx→0 2 / (x2 - 16)
= 2 / (0 - 16)
= -1/8
Wiki User
∙ 12y ago
Add your answer:
Earn +20 pts
Solve x over the quantity x minus one equals onehalf minus the quantity x plus one over the quantity x minus two?
irdk
How do you solve 8 minus n over 7 equals 4 over 3?
n=-4/3
Why the number 0 has no reciprocal?
Division by zero is not allowed/defined. So you cannot take 'one over zero', or have zero in the denominator.Without going too technical, a person might say that 1/0 is infinity, and it sounds good. But if you have a function [say f(x) = 1/x] and take the limit of f(x) as x approaches zero, then f(x) approaches infinity as x approaches from the right, but it approaches negative infinity as you approach from the left, therefore the limit does not exist.
What is 6 over 7 minus 5 over 7?
6 over 7 minus 5 over 7 is 1 over 7.
What is the answer to 13 over 36 minus 9 over 35?
The answer to 13/36 minus 9/35 is 131/1260
What is the limit of sin multiplied by x minus 1 over x squared plus 2 as x approaches infinity?
As X approaches infinity it approaches close as you like to 0. so, sin(-1/2)
What is the limit of x squared plus four x plus four all over x squared minus 4 as x approaches 2?
The "value" of the function at x = 2 is (x+2)/(x-2) so the answer is plus or minus infinity depending on whether x approaches 2 from >2 or <2, respectively.
What is the limit of x cosine 1 over x squared as x approaches 0?
The limit is 0.
What is the limit of sine squared x over x as x approaches zero?
So, we want the limit of (sin2(x))/x as x approaches 0. We can use L'Hopital's Rule: If you haven't learned derivatives yet, please send me a message and I will both provide you with a different way to solve this problem and teach you derivatives! Using L'Hopital's Rule yields: the limit of (sin2(x))/x as x approaches 0=the limit of (2sinxcosx)/1 as x approaches zero. Plugging in, we, get that the limit is 2sin(0)cos(0)/1=2(0)(1)=0. So the original limit in question is zero.
What is the Limits of h over tan h as h approaches 0?
The limit is 1.
How do you solve 2 minus five over seven?
2 minus 5 is -3, sooo your answer should be negative 3 over 7.
Solve x over the quantity x minus one equals onehalf minus the quantity x plus one over the quantity x minus two?
irdk
How do you solve 2u square minus 14u over 5u square times 15u square over 4u minus 28?
3u2-1
How do you solve 8 minus n over 7 equals 4 over 3?
n=-4/3
What is the limit as x approaches 0 of 1 plus cos squared x over cos x?
2
How do you solve eight minus two and three fourths?
turn 8 into 8 over 8 and solve from there.
How do you solve x over x minus 2 minus 3 over 2?
x over x is one, so the problem would be 1-2-3/2=-5/3
