The limit in question is written in a somewhat complex form, but it can be simplified and solved using 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.

The limit is: lim (x→0+) [x^(-1)sin(8x)]

This is a 0/0 indeterminate form, so we can apply L'Hopital's rule.

First, let's find the derivatives of the numerator and the denominator:

The derivative of x^(-1) is -x^(-2) and the derivative of sin(8x) is 8cos(8x).

So, the limit becomes: lim (x→0+) [-x^(-2)/8cos(8x)]

This is still a 0/0 indeterminate form, so we can apply L'Hopital's rule again.

The derivative of -x^(-2) is 2x^(-3) and the derivative of 8cos(8x) is -64sin(8x).

So, the limit becomes: lim (x→0+) [-2x^(-3)/-64sin(8x)]

This simplifies to: lim (x→0+) [x^3/32sin(8x)]

As x approaches 0, sin(8x) approaches 0 and x^3 also approaches 0. Therefore, the limit is 0.

So, the correct answer is D. limit as x rightwards arrow 0 to the power of plus of x to the power of short dash 1 end exponent sin left parenthesis 8 x right parenthesis equals 0.

Question

The limit in question is written in a somewhat complex form, but it can be simplified and solved using 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.

The limit is: lim (x→0+) [x^(-1)sin(8x)]

This is a 0/0 indeterminate form, so we can apply L'Hopital's rule.

First, let's find the derivatives of the numerator and the denominator:

The derivative of x^(-1) is -x^(-2) and the derivative of sin(8x) is 8cos(8x).

So, the limit becomes: lim (x→0+) [-x^(-2)/8cos(8x)]

This is still a 0/0 indeterminate form, so we can apply L'Hopital's rule again.

The derivative of -x^(-2) is 2x^(-3) and the derivative of 8cos(8x) is -64sin(8x).

So, the limit becomes: lim (x→0+) [-2x^(-3)/-64sin(8x)]

This simplifies to: lim (x→0+) [x^3/32sin(8x)]

As x approaches 0, sin(8x) approaches 0 and x^3 also approaches 0. Therefore, the limit is 0.

So, the correct answer is D. limit as x rightwards arrow 0 to the power of plus of x to the power of short dash 1 end exponent sin left parenthesis 8 x right parenthesis equals 0.

Knowee AI · Accepted Answer