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
irdk
n=-4/3
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.
6 over 7 minus 5 over 7 is 1 over 7.
The answer to 13/36 minus 9/35 is 131/1260
As X approaches infinity it approaches close as you like to 0. so, sin(-1/2)
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.
The limit is 0.
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.
2 minus 5 is -3, sooo your answer should be negative 3 over 7.
The limit is 1.
irdk
3u2-1
n=-4/3
turn 8 into 8 over 8 and solve from there.
x over x is one, so the problem would be 1-2-3/2=-5/3
2