if
f(x)=x[x]/sin[x]
then show that lim f(x)has no existence
x>0
assuming you are using the [..] operator to have the same meaning as (..)
we have
Limit x^2/sinx , as x ---> 0
using L'Hopitals Rule
lim x^2/sinx
= lim 2x/cosx , as x --->
= 0/1
= 0
verification:
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2%2Fsinx
looks at x = 0,
the function has a value of 0 , the origin (0,0)
If [ ... ] is some other mathematical operator, define it please
as an example....
[2.32]=2
To show that the limit of f(x) does not exist as x tends to a positive value, we can use the concept of oscillation.
First, let's analyze the given function f(x) = (x[x])/sin[x]:
The expression x[x] represents the greatest integer less than or equal to x. For example, [2.5] = 2 and [3.8] = 3.
The denominator sin[x] is a periodic function with infinitely many oscillations as x approaches positive values.
To show that the limit of f(x) does not exist, we need to demonstrate that the function oscillates between different values as x approaches the designated positive value.
Let's consider an example:
Suppose we examine the behavior of f(x) as x approaches 2 from the right side (x > 0).
As x approaches 2, the expression x[x] evaluates to 2[2] = 2*2 = 4.
The denominator sin[x] also oscillates as x approaches 2.
When x = 2, sin[x] = sin[2].
Now, if we consider values slightly greater than 2, like x = 2.1, 2.01, 2.001, and so on, the numerator x[x] will remain 4 since we're still between 2 and 3.
However, the denominator sin[x] will oscillate between different values as we approach 2. It can take values like sin[2.1], sin[2.01], sin[2.001], and so on, which will be different from sin[2] but still approach zero.
Hence, the function f(x) = (x[x])/sin[x] will oscillate between different values as x approaches a positive value. Consequently, the limit of f(x) will not exist for x > 0.