if

f(x)=x[x] / sin[x]
then show that there is no existence of

lim f(x)
x>0

2nd prob:

show lim (tanx-sinx)/x^3 =1/2
x>0

To show that the limit of a function exists, we need to prove that the left-hand limit and the right-hand limit are equal. Let's tackle each problem one at a time.

For the first problem:

We want to evaluate the limit:
lim f(x) as x approaches 0+ (from the right).

To do this, we need to find the left-hand limit and the right-hand limit first.

For the right-hand limit, where x approaches 0+, we can substitute x directly into the function and simplify:
lim (x * [x] / sin[x]) as x approaches 0+

Since [x] represents the greatest integer less than or equal to x, we can rewrite the function as:
lim (x * floor(x) / sin[x]) as x approaches 0+

Here, the numerator (x * floor(x)) approaches 0 as x approaches 0+ because both x and floor(x) approach 0. The denominator (sin[x]) also approaches 0 as x approaches 0 because sin[x] is continuous at x = 0. Therefore, we have an indeterminate form of 0/0.

To evaluate this, we can use L'Hôpital's Rule. Differentiating the numerator and the denominator, we get:
lim (floor(x) + x * d(floor(x))/dx) / d(sin[x])/dx) as x approaches 0+

The derivative of floor(x) is 0 since it is a step function and its derivative is defined to be zero. The derivative of sin[x] is cos[x].

Simplifying further, we have:
lim (x * 0 + 1 * cos[x]) / cos[x] as x approaches 0+
lim (cos[x]) / cos[x] as x approaches 0+
lim 1 as x approaches 0+
1

Therefore, the right-hand limit of f(x) as x approaches 0+ is 1.

Next, let's evaluate the left-hand limit.

For the left-hand limit, we need to evaluate:
lim f(x) as x approaches 0- (from the left).

However, when we examine the given function, we notice that the denominator, sin[x], changes signs infinitely close to x = 0, making the function undefined as x approaches 0-.

Since the left-hand limit does not exist, the limit of f(x) as x approaches 0 from both sides cannot be the same. Therefore, there is no existence of the limit lim f(x) as x approaches 0.

Moving on to the second problem:

We want to evaluate the limit:
lim ((tan x - sin x) / x^3) as x approaches 0+.

To solve this limit, we can use Taylor series expansion. The Taylor series expansion of the tangent function is:
tan x = x + (1/3)x^3 + (2/15)x^5 + ...

The Taylor series expansion of the sine function is:
sin x = x - (1/6)x^3 + (1/120)x^5 - ...

Now, let's substitute these expansions into the original limit expression:
lim ((x + (1/3)x^3 + ...) - (x - (1/6)x^3 + ...)) / x^3 as x approaches 0+

Simplifying, we get:
lim ((2/3)x^3 + ...) / x^3 as x approaches 0+
lim (2/3 + ...) as x approaches 0+
2/3

Therefore, the limit of ((tan x - sin x) / x^3) as x approaches 0 from the right is 2/3.

Since the limit from the right exists and is finite, we can conclude that the limit of ((tan x - sin x) / x^3) as x approaches 0 exists and is equal to 2/3.

Note that we have shown the limit exists as x approaches 0 from the right (x > 0), whereas your question states x > 0 without specifying the direction of approach.