0
Undefined: You cannot divide by zero
Alex Kuhic
Wiki User
∙ 11y agoAs x approaches 0, The form of the equation becomes '0/0.'
This is an indeterminate form. Using L'Hopitals Rules, you get
if y = cosx/x, y = -sinx/1. This, as x tends to 0 becomes -1.
Add your answer:
Earn +20 pts
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.
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.
What is limit as x approaches 0 of sin squared x by x?
lim(x->0) of sin(x)^2/x we use L'Hospital's Rule and derive the top and the bottomd/dx(sin(x)^2/x)=2*sin(x)*cos(x)/1lim(x->0) of 2*sin(x)*cos(x)=2*0*1=0
What is the limit as x approaches zero of 5x divided by 3sinxcosx?
1.6667
If f(x) cos(2x).cos(4x).cos(6x).cos(8x).cos(10x) then find the limit of 1 - f(x)35(sinx)2 as x tends to 0 (zero).?
I'm sorry the question is not correctly displayed. If f(x) = cos(2x).cos(4x).cos(6x).cos(8x).cos(10x) then, find the limit of {1 - [f(x)]^3}/[5(sinx)^2] as x tends to 0 (zero).
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 of 1 -cos x divided by x as x approaches 0?
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.
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 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.
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.
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.
