Limit (TIFR 2013 problem 12)



$$ \lim_{x\to 0} \frac{sin(x^2)}{x^2}sin(\frac{1}{x}) = 1$$


As \(x\) goes to zero, \(x^2\) also goes to zero, and consequently, since $$ \lim_{y\to 0} \frac{sin(y)}{y}=1 $$ we have $$ \lim_{x\to 0} \frac{sin(x^2)}{x^2}= 1$$. If the statement were true, then \(\lim_{x\to 0}sin(\frac{1}{x}) = 1\), which is false. In fact, \(\lim_{x\to 0}sin(\frac{1}{x}) \) does not exist.

So the statement is False. 

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