Calculate the following limits:(a) limx→0sin(2x2 + x3)x ; (b) limx→0√x(sin(1/x) − cos(1/x)).

(a) limx→0sin(2x2 + x3)/x

Step 1: We can see that direct substitution will give us 0/0 which is an indeterminate form. So, 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.

Step 2: Differentiate the numerator and the denominator.

The derivative of sin(2x2 + x3) with respect to x is (4x + 3x2)cos(2x2 + x3) using the chain rule.

The derivative of x with respect to x is 1.

Step 3: Substitute these into our limit, we get:

limx→0 (4x + 3x2)cos(2x2 + x3)

Step 4: Now, substitute x = 0 into the limit, we get:

(40 + 30^2)cos(2*0^2 + 0^3) = 0

So, limx→0sin(2x2 + x3)/x = 0

(b) limx→0√x(sin(1/x) − cos(1/x))

Step 1: Again, direct substitution will give us an indeterminate form. But in this case, L'Hopital's Rule is not applicable because we don't have a quotient of two functions.

Step 2: We can rewrite the limit as:

limx→0 √x * sin(1/x) - limx→0 √x * cos(1/x)

Step 3: Now, we can see that both of these limits are in the form 0*∞ which is an indeterminate form. We can convert this to the form 0/0 or ∞/∞ by taking the reciprocal of one of the factors.

Step 4: For the first limit, we can rewrite it as:

limx→0 (sin(1/x))/(1/√x)

Now, we can apply L'Hopital's Rule. The derivative of sin(1/x) with respect to x is -(1/x^2)cos(1/x) and the derivative of 1/√x with respect to x is -(1/2x^3/2).

Substitute these into the limit, we get:

limx→0 [-(1/x^2)cos(1/x)] / [-(1/2x^3/2)]

Simplify this, we get:

limx→0 2x^1/2 cos(1/x)

Now, substitute x = 0 into the limit, we get:

2*0^1/2 cos(1/0) = 0

Step 5: For the second limit, we can rewrite it as:

limx→0 (cos(1/x))/(1/√x)

Now, we can apply L'Hopital's Rule. The derivative of cos(1/x) with respect to x is (1/x^2)sin(1/x) and the derivative of 1/√x with respect to x is -(1/2x^3/2).

Substitute these into the limit, we get:

limx→0 [(1/x^2)sin(1/x)] / [-(1/2x^3/2)]

Simplify this, we get:

limx→0 -2x^1/2 sin(1/x)

Now, substitute x = 0 into the limit, we get:

-2*0^1/2 sin(1/0) = 0

Step 6: So, the original limit is 0 - 0 = 0.

So, limx→0√x(sin(1/x) − cos(1/x)) = 0.