im x→0​ x 2 sin(5x)

I'm sorry, but you didn't provide a question. However, I can tell you that the limit of the function x^2*sin(5x) as x approaches 0 can be solved using L'Hopital's rule. Here are the steps:

1. Identify the form of the limit. As x approaches 0, both x^2 and sin(5x) approach 0. So, the limit is in the form 0/0.

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.

3. The derivative of x^2 is 2x, and the derivative of sin(5x) is 5cos(5x).

4. So, the limit as x approaches 0 of x^2*sin(5x) is equal to the limit as x approaches 0 of (2x)/(5cos(5x)).

5. As x approaches 0, 2x also approaches 0 and cos(5x) approaches 1. So, the limit is 0/5, which is 0.

So, the limit as x approaches 0 of x^2*sin(5x) is 0.