Calculate the following limits:(i) limx→0sin(x2)x ; (ii) limx→0 x2 sin(1/x).Hint) You may use the properties of trigonometric functions in the paragraph below Ex-ample 5.9

(i) limx→0sin(x2)/x

This limit is in an indeterminate form 0/0. We can use L'Hopital's rule, which states that the limit of a quotient of two functions as x approaches a certain value is equal to the limit of the quotients of their derivatives.

First, we need to find the derivatives of the numerator and the denominator.

The derivative of sin(x^2) with respect to x is 2x cos(x^2) using the chain rule.

The derivative of x is 1.

So, we have:

limx→0sin(x2)/x = limx→0(2x cos(x^2))/1

As x approaches 0, 2x cos(x^2) also approaches 0. Therefore, the limit is 0.

(ii) limx→0 x^2 sin(1/x)

This limit is also in an indeterminate form 0*infinity. We can rewrite the expression as a quotient and then apply L'Hopital's rule.

Rewrite x^2 sin(1/x) as (x^2)/ (1/sin(1/x)).

Now, we need to find the derivatives of the numerator and the denominator.

The derivative of x^2 with respect to x is 2x.

The derivative of 1/sin(1/x) with respect to x is -cos(1/x)/x^2 using the chain rule.

So, we have:

limx→0 x^2 sin(1/x) = limx→0 (2x)/(-cos(1/x)/x^2)

As x approaches 0, 2x also approaches 0 and -cos(1/x)/x^2 approaches infinity. Therefore, the limit is 0.

(i) limx→0sin(x2)/x

To solve this limit, we can use the squeeze theorem. The squeeze theorem states that if we have three functions, f(x), g(x), and h(x), and if f(x) ≤ g(x) ≤ h(x) for all x in an interval around a, except possibly at a itself, and if limx→a f(x) = limx→a h(x) = L, then limx→a g(x) = L.

In this case, we can use the fact that -1 ≤ sin(x) ≤ 1 for all x. Therefore, -x^2 ≤ sin(x^2) ≤ x^2 for all x. Dividing through by x, we get -x ≤ sin(x^2)/x ≤ x. As x approaches 0, both -x and x approach 0, so by the squeeze theorem, limx→0 sin(x^2)/x = 0.

(ii) limx→0 x^2 sin(1/x)

This limit can be solved directly. As x approaches 0, x^2 approaches 0. The limit of sin(1/x) as x approaches 0 does not exist, but it is bounded between -1 and 1. Therefore, the product x^2 sin(1/x) is also bounded between -x^2 and x^2. As x approaches 0, both -x^2 and x^2 approach 0, so by the squeeze theorem, limx→0 x^2 sin(1/x) = 0.