0
The limit is 1.
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How do you solve the limit as x approaches 90 degrees of cos 2x divided by tan 3x?
Take the limit of the top and the limit of the bottom. The limit as x approaches cos(2*90°) is cos(180°), which is -1. Now, take the limit as x approaches 90° of tan(3x). You might need a graph of tan(x) to see the limit. The limit as x approaches tan(3*90°) = the limit as x approaches tan(270°). This limit does not exist, so we'll need to take the limit from each side. The limit from the left is ∞, and the limit from the right is -∞. Putting the top and bottom limits back together results in the limit from the left as x approaches 90° of cos(2x)/tan(3x) being -1/∞, and the limit from the right being -1/-∞. -1 divided by a infinitely large number is 0, so the limit from the left is 0. -1 divided by an infinitely large negative number is also zero, so the limit from the right is also 0. Since the limits from the left and right match and are both 0, the limit as x approaches 90° of cos(2x)/tan(3x) is 0.
What is the limit of x cosine 1 over x squared as x approaches 0?
The limit is 0.
If for a triangle abc tan a-b plus tan b-c plus tan c-a equals 0 then what can you say about the triangle?
tan (A-B) + tan (B-C) + tan (C-A)=0 tan (A-B) + tan (B-C) - tan (A-C)=0 tan (A-B) + tan (B-C) = tan (A-C) (A-B) + (B-C) = A-C So we can solve tan (A-B) + tan (B-C) = tan (A-C) by first solving tan x + tan y = tan (x+y) and then substituting x = A-B and y = B-C. tan (x+y) = (tan x + tan y)/(1 - tan x tan y) So tan x + tan y = (tan x + tan y)/(1 - tan x tan y) (tan x + tan y)tan x tan y = 0 So, tan x = 0 or tan y = 0 or tan x = - tan y tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = - tan(B-C) tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = tan(C-B) A, B and C are all angles of a triangle, so are all in the range (0, pi). So A-B and B-C are in the range (- pi, pi). At this point I sketched a graph of y = tan x (- pi < x < pi) By inspection I can see that: A-B = 0 or B-C = 0 or A-B = C-B or A-B = C-B +/- pi A = B or B = C or A = C or A = C +/- pi But A and C are both in the range (0, pi) so A = C +/- pi has no solution So A = B or B = C or A = C A triangle ABC has the property that tan (A-B) + tan (B-C) + tan (C-A)=0 if and only if it is isosceles (or equilateral).
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)
When does a problem in mathematics have no limit?
When the limit as the function approaches from the left, doesn't equal the limit as the function approaches from the right. For example, let's look at the function 1/x as x approaches 0. As it approaches 0 from the left, it travels towards negative infinity. As it approaches 0 from the right, it travels towards positive infinity. Therefore, the limit of the function as it approaches 0 does not exist.
How do you solve the limit as x approaches 90 degrees of cos 2x divided by tan 3x?
Take the limit of the top and the limit of the bottom. The limit as x approaches cos(2*90°) is cos(180°), which is -1. Now, take the limit as x approaches 90° of tan(3x). You might need a graph of tan(x) to see the limit. The limit as x approaches tan(3*90°) = the limit as x approaches tan(270°). This limit does not exist, so we'll need to take the limit from each side. The limit from the left is ∞, and the limit from the right is -∞. Putting the top and bottom limits back together results in the limit from the left as x approaches 90° of cos(2x)/tan(3x) being -1/∞, and the limit from the right being -1/-∞. -1 divided by a infinitely large number is 0, so the limit from the left is 0. -1 divided by an infinitely large negative number is also zero, so the limit from the right is also 0. Since the limits from the left and right match and are both 0, the limit as x approaches 90° of cos(2x)/tan(3x) is 0.
What is the limit as t approaches 0 of 2t tan t?
That would be 0. (But that was too easy. Did you mean something else?)
What is sin7x divided by tan4x?
1.75
How do you make the left the same as the right for tan 0 times cot 0 all over tan 0 csc2 0?
You cannot because cot(0) is not defined, neither is cosec(0).
What is the limit of x cosine 1 over x squared as x approaches 0?
The limit is 0.
What is the exact trigonometric function value of cot 0?
There is no value cot 0, because cot 0 is equivalent to 1 / tan 0, which is equivalent to 1 / 0, which is undefined. That said, the limit of cot x as x approaches 0 is infinity.
What is tan of 0?
Tan of 0 equals zero.
What is tan of 90 degrees?
On the unit circle at 90 degrees the 90 degrees in radians is pi/2 and the coordinates for this are: (0,1). The tan function = sin/cos. In the coordinate system x is cos and y is sin. Therefore (0,1) ; cos=0, & sin=1 . Tan=sin/cos so tan of 90 degrees = 1/0. The answer of tan(90) = undefined. There can not be a 0 in the denominator, because you can't devide by something with no quantity. Something with no quantity is 0. Or, on a limits point of view, it would be infinity.
Lim x approaches 0 x x x x?
When the limit of x approaches 0 x approaches the value of x approaches infinity.
If for a triangle abc tan a-b plus tan b-c plus tan c-a equals 0 then what can you say about the triangle?
tan (A-B) + tan (B-C) + tan (C-A)=0 tan (A-B) + tan (B-C) - tan (A-C)=0 tan (A-B) + tan (B-C) = tan (A-C) (A-B) + (B-C) = A-C So we can solve tan (A-B) + tan (B-C) = tan (A-C) by first solving tan x + tan y = tan (x+y) and then substituting x = A-B and y = B-C. tan (x+y) = (tan x + tan y)/(1 - tan x tan y) So tan x + tan y = (tan x + tan y)/(1 - tan x tan y) (tan x + tan y)tan x tan y = 0 So, tan x = 0 or tan y = 0 or tan x = - tan y tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = - tan(B-C) tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = tan(C-B) A, B and C are all angles of a triangle, so are all in the range (0, pi). So A-B and B-C are in the range (- pi, pi). At this point I sketched a graph of y = tan x (- pi < x < pi) By inspection I can see that: A-B = 0 or B-C = 0 or A-B = C-B or A-B = C-B +/- pi A = B or B = C or A = C or A = C +/- pi But A and C are both in the range (0, pi) so A = C +/- pi has no solution So A = B or B = C or A = C A triangle ABC has the property that tan (A-B) + tan (B-C) + tan (C-A)=0 if and only if it is isosceles (or equilateral).
If tansqtheta plus 5tantheta0 find the value of tantheta plus cottheta?
tan2(theta) + 5*tan(theta) = 0 => tan(theta)*[tan(theta) + 5] = 0=> tan(theta) = 0 or tan(theta) = -5If tan(theta) = 0 then tan(theta) + cot(theta) is not defined.If tan(theta) = -5 then tan(theta) + cot(theta) = -5 - 1/5 = -5.2
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)
