Wiki User
∙ 14y agoThe limit doesn't exist. Consider approaching from the left (->) and the right (<-) [I'm assuming x is real, but if it is complex the same holds].
(->) : (x-1) / |x-1|. Since x<1, |x-1| = 1-x. So, we have
(x-1) / (1-x) = (x-1) / -1(x-1) = -1
(<-) : (x-1) / |x-1|. Since x>1, |x-1| = x-1. So, we have
(x-1) / (x-1) = 1
But recall that limits must be unique and clearly 1 doesn't equal -1. Thus, the limit doesn't exist.
Wiki User
∙ 14y agoThat's 1/2 .(You have to use l'Hospital's rule.)
It can be difficult to remember all mathematical terms and their meanings. The limit concept is the value that a function or sequence approaches as the input approaches a value.Ê
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.
The absolute value of 3 divided by 15 is 0.2.
350.
That's 1/2 .(You have to use l'Hospital's rule.)
A limit is the value that a function approaches as the input gets closer to a specific value.
When the limit of x approaches 0 x approaches the value of x approaches infinity.
It can be difficult to remember all mathematical terms and their meanings. The limit concept is the value that a function or sequence approaches as the input approaches a value.Ê
The Third Law of Thermodynamics states that absolute zero cannot be reached. This law asserts that as a system approaches absolute zero, its entropy approaches a minimum value but never reaches zero.
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.
No, limit can tend to any finite number including 0. It is also possible that the limit does not tend to any finite value or approaches infinity. Example: The limit of x^2+5 tend to 6 as x approaches -1.
The absolute value of 3 divided by 15 is 0.2.
350.
To find the uncertainty when a constant is divided by a value with an uncertainty, you can use the formula for relative uncertainty. Divide the absolute uncertainty of the constant by the value, and add it to the absolute uncertainty of the value divided by the value squared. This will give you the combined relative uncertainty of the division.
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.
For any value divided by zero the answer is "infinite" or "undefined". Think about it. X divided by 1 is X. X divided by 1/2 is 2X. X divided by 1/4 is 4X. As the divisor gets smaller, the result gets larger. As the divisor approaches zero, the result approaches infinity.