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What is limit of 1 -cos x divided by x as x approaches 0?
1
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 does a number over infinity equal?
0. This is the same as the limit of 1/x as x approaches infinity, which is is 0. This is because 1/1,000 = .001 and 1/1,000,000 = .000001 and 1/100,000,000,000 = .0000000001 etc.
What is the 2 divided by infinity?
It is 0. Think of dividing 2 by a very big number. For example, 2/20000000000= 1/10000000000 which is very close to 0. As the denominator gets bigger and bigger, the quotient approaches 0. In the limit it is 0. Remember infinity is not really a number. In this case it means, letting the denominator get as big as it can. Another way to think of or write this this is lim n-->0 of 2/n=0
How do I evaluate the limit sin9x divided by tan9x when x tends to 0 without using L Hospitals rule?
First to simplify matters, change y=9x. So we are looking at limit sin(y) divided by tan(y).Now lets look at right angled triangle wheresin(y) = a/ctan(y) = a/bthus we are looking at the limit of (a/c)/(a/b) = limit of b/cAs the angle y shrinks, the right angle remains constant, and the remaining angle approaches a right angle. Thus at the limit we have a triangle with equal angles and thus where b=c.As a result limit you are trying to calculate is 1.
What is limit as x approaches 0 of cos squared x by x?
The limit of cos2(x)/x as x approaches 0 does not exist. As x approaches 0 from the left, the limit is negative infinity. As x approaches 0 from the right, the limit is positive infinity. These two values would have to be equal for a limit to exist.
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.
Is zero divided by zero equal to zero?
Actually 0/0 is undefined because there is no logical way to define it. In ordinary mathematics, you cannot divide by zero.The limit of x/x as x approaches 0 exists and equals 1 so you might be tempted to define 0/0 to be 1.However, the limit of x2/x as x approaches 0 is 0, and the limit of x/x2 as x approaches 0 does not exist .r/0 where r is not 0 is also undefined. It is certainly misleading, if not incorrect to say that r/0 = infinity.If r > 0 then the limit of r/x as x approaches 0 from the right is plus infinity which means the expression increases without bounds. However, the limit as x approaches 0 from the left is minus infinity.
Is it true that infinity goes both ways?
Yes, ∞ and -∞ both exist as distinct entities. If I take the limit of 1/x as x approaches 0 from the positive side, I get ∞. On the other hand if I take the limit of 1/x as x approaches 0 from the negative side, I get -∞.
What is 0 to the power of 0?
00 is undefined.Consider the limit of x0 as x approaches 0. Since x0 = 1 for every positive value of x, the limit is 1.Now consider the limit of 0x as xapproaches 0. This limit is 0. Since the two limits are unequal, the value of 00 cannot exist.
Lim x approaches 0 x x x x?
When the limit of x approaches 0 x approaches the value of x approaches infinity.
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.
Can a chord also be a tangent?
In the limit, a chord approaches a tangent, but is never actually a tangent. (In much the same way as 1/x approaches 0 as x increases, but is never actually 0.)In the limit, a chord approaches a tangent, but is never actually a tangent. (In much the same way as 1/x approaches 0 as x increases, but is never actually 0.)In the limit, a chord approaches a tangent, but is never actually a tangent. (In much the same way as 1/x approaches 0 as x increases, but is never actually 0.)In the limit, a chord approaches a tangent, but is never actually a tangent. (In much the same way as 1/x approaches 0 as x increases, but is never actually 0.)
What is the limit of x2 as x approaches 0?
Here, just plug x=0 into x^2 to get 0^2=0. The limit is 0.
Why absolute value of x is not differentiable at point 0?
The absolute value of x, |x|, is defined as |x| = x, x>=0; -x, x<0. If you derive this, then you will find that the derivative is 1 when x>=0, and -1 when x<0. But this means that the derivative as x approaches 0 from the left does not equal the derivative as x approaches 0 from the right, as -1=/=1. So the limit as x approaches 0 does not exist, and therefore the gradient does not exist at that point, and so |x| cannot be differentiated at x = 0.
Lim x approaches 0 x x x x-?
When the limit of x approaches 0 the degree on n is greater than 0.
