The 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.
That'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 absolute value of 3 divided by 15 is 0.2.
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.
350.
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.
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.
Because infinity is not a umber, it is usually not treated as a number when computing functions. Instead, you can look for a limit of a function as it approaches infinity. For example, the limit as x approaches infinity of 1/x is 0. Because sine oscillates, it's value constantly moves up and down, and it's value as it approaches infinity is not defined because it does not converge on any one number, as some other functions (like 1/x) do.
That is called the "absolute value" of the number. For example:The absolute value of 5 is 5.The absolute value of -5 is also 5.That is called the "absolute value" of the number. For example:The absolute value of 5 is 5.The absolute value of -5 is also 5.That is called the "absolute value" of the number. For example:The absolute value of 5 is 5.The absolute value of -5 is also 5.That is called the "absolute value" of the number. For example:The absolute value of 5 is 5.The absolute value of -5 is also 5.