Evaluate the limit: lim (x -> 0) (x^2 * sin(1/x))

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Question: Evaluate the limit: lim (x -> 0) (x^2 * sin(1/x))

Options:

Correct Answer: 0

Solution:

Since |sin(1/x)| <= 1, we have |x^2 * sin(1/x)| <= x^2, and thus lim (x -> 0) x^2 * sin(1/x) = 0.

Evaluate the limit: lim (x -> 0) (x^2 * sin(1/x))

Evaluate the limit: lim (x -> 0) (x^2 * sin(1/x))

Correct Answer: 0

Step 1: Understand the limit we want to evaluate: lim (x -> 0) (x^2 * sin(1/x)).
Step 2: Recall that the sine function, sin(y), always gives values between -1 and 1 for any real number y.
Step 3: Since 1/x becomes very large as x approaches 0, we know that sin(1/x) will oscillate between -1 and 1.
Step 4: This means that |sin(1/x)| is always less than or equal to 1, so we can write: |sin(1/x)| <= 1.
Step 5: Multiply both sides of the inequality by x^2 (which is always non-negative): |x^2 * sin(1/x)| <= x^2.
Step 6: Now we can analyze the limit of x^2 as x approaches 0: lim (x -> 0) x^2 = 0.
Step 7: Since |x^2 * sin(1/x)| is less than or equal to x^2, and x^2 approaches 0, we can conclude that lim (x -> 0) x^2 * sin(1/x) = 0.

Limit Evaluation – Understanding how to evaluate limits involving oscillatory functions and applying the squeeze theorem.
Squeeze Theorem – Using the properties of bounded functions to determine the limit of a product involving a bounded oscillatory function.