Sorry, there seems to be a misunderstanding. I believe you want to know how to solve the limit problem: limx→0+(tan (2x))x. Here are the steps:

Step 1: Recognize that this is an indeterminate form of type 0/0 since as x approaches 0, both tan(2x) and x approach 0.

Step 2: Apply 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 3: Differentiate the numerator and the denominator. The derivative of tan(2x) is 2sec^2(2x) and the derivative of x

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Sorry, there seems to be a misunderstanding. I believe you want to know how to solve the limit problem: limx→0+(tan (2x))x. Here are the steps:

Step 1: Recognize that this is an indeterminate form of type 0/0 since as x approaches 0, both tan(2x) and x approach 0.

Step 2: Apply 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 3: Differentiate the numerator and the denominator. The derivative of tan(2x) is 2sec^2(2x) and the derivative of x

Knowee AI · Accepted Answer

Sorry, there seems to be a misunderstanding. I believe you want to know how to solve the limit problem: limx→0+(tan (2x))x. Here are the steps:

Step 1: Recognize that this is an indeterminate form of type 0/0 since as x approaches 0, both tan(2x) and x approach 0.

Step 2: Apply 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 3: Differentiate the numerator and the denominator. The derivative of tan(2x) is 2sec^2(2x) and the derivative of x

limx→0+(tan (2x))x

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