MHB Double-Check My Problem Solving | Correct Reasoning Verification

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The discussion centers on verifying the correctness of a problem-solving approach regarding a continuous function at x=0. The initial poster acknowledges a misunderstanding of the question, clarifying that while the function is continuous, demonstrating that the limit as x approaches 0 equals 1 is necessary. Albert's reasoning is challenged, specifically regarding the inequalities he presented, which are incorrect near x=0. Additionally, there are errors in the limits he calculated, as they do not equal 1 but should be 0 instead. The conversation emphasizes the importance of accurately applying mathematical principles in problem-solving.
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I am not entirely sure if I solved this problem correctly. Please let me know if my reasoning is flawed. Thank you and I appreciate your help greatly.
 

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The function is continuous, just substitute x = 0...
 
On second thought, I didn't read the question properly... Please disregard my previous post.
 
The function IS continuous at x=0 but you need to show that $\lim_{x\to 0} f(x)= 1$ to show THAT so continuity cannot be used to do this exercise.

Albert, you state that $-1\le x^2- 2\le 1$. That is the same as $1\le x^2\le 3$. If x is close to 0 that is NOT true! You also have $\lim_{x\to 0} -x^2= 1$ and $\lim_{x\to 0} x^2= 1$. Those are certainly not true! Did you mean "= 0"?
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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