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Ophelia Schneider
Wiki User
∙ 13y agoThe limit of a fraction is equivalent to the limit of the derivative of the numerator divided by the limit of the denominator, so
lim ((1-cos(x)) / x) as x -> 0 is the same as
lim (sin(x) / 1) as x -> 0
Therefore as x tends to zero, the function tends to 0
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What is the lim of h if it equals 0 Sinxcosh plus cosxsinh minus sinx divided by h?
lim(h→0) (sin x cos h + cos x sin h - sin x)/h As h tends to 0, both the numerator and the denominator have limit zero. Thus, the quotient is indeterminate at 0 and of the form 0/0. Therefore, we apply l'Hopital's Rule and the limit equals: lim(h→0) (sin x cos h + cos x sin h - sin x)/h = lim(h→0) (sin x cos h + cos x sin h - sin x)'/h' = lim(h→0) [[(cos x)(cos h) + (sin x)(-sin h)] + [(-sin x)(sin h) + (cos x)(cos h)] - cos x]]/0 = cosx/0 = ∞
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
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.
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 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.
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 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.
What is the limit as x approaches 0 of 1 plus cos squared x over cos x?
2
What is the limit of 1-cos x over x squared when x approaches 0?
1
How do you calculate the limit of e5x -1 divided by sin x as x approaches 0?
You can use the L'hopital's rule to calculate the limit of e5x -1 divided by sin x as x approaches 0.
What is limit of 1 cos x divided by x as x approaches 0?
Undefined: You cannot divide by zero
What is the value of 1- cos x when x is small?
1 - cos x as x approaches 0. what is the cos of 0? It is 1. So as x approaches 0 cos x approaches 1. 1 - 1 = 0 So as it gets very small the solutions gets smaller.
How do you evaluate this limit Algebraically limit as x approaches 0 of 3sinx divided by x-2tanx without using Lhospital rule?
sinx = sin0 = 0 tanx = tan0 = 0 you have 0/0 by you limit conditions
What is the limit of 9x divided by 2x as x approaches 0?
9x/2x = 9/2 = 4.5
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 lim as x approaches 0 of 3 times 1 - cosx divided by x?
First, you can take the constant factor 3 out, to obtain 3 times the limit of (1 - cos x) / x. Since this is of the form 0/0, you can use L'Hôpital's rule, which states that in such cases, you can take the derivative of both numerator and denominator. This results in the limit (as x approaches 0) of sin x / x, that is, 1 / 1 = 1. So, the final result is 3 times the limit of 1 = 3.
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.
