<p>Ok, for all you BC Calc geniuses out there, this problem involves the Squeeze Theorum (Sandwhich Theorum)
Find the limit as x approaches 0 of (X^2)(sinx)
Be sure to do it using an inequality and absolute value. My question is how you can infer the original function from the absolute value of the function.</p>
<p>bumppppppp</p>
<p>|sin x|<=1
call your function f(x)
let g(x)=x^2
let h(x)=-x^2
then h(x)<=f(x)<=g(x) (on any interval, but pick (-1,1))
obviously, lim x-->0 g(x) = lim x-->0 h(x) = 0 (both are polynomials, so are continuous so the limit of each at zero is just its value at 0)
so lim x-->0 f(x) = 0</p>
<p>QED</p>
<p>Not bad for someone who took calculus in 1973, eh?</p>
<p>PS -</p>
<p>No more help with your homework!</p>
<p>yes i got that far...but the problem is you would still have
0<= lim as x-->0 of |x^2 sinx|<=0
how do you prove that this is the same as
0<= lim as x-->0 of x^2 sinx<=0 (no absolute value)</p>
<p>bumpppppppppppp</p>
<p>If you're having trouble I'd suggest asking Mr. Donoughe.</p>
<p>dolphins -</p>
<p>You don't need to prove that, all you need to prove is that f (which is (sinx)*(x^2); absolute value doesn't enter in at that point) is between g and h on an open interval around 0 (but not necessarily including 0 itself). If that's the case and if g and h converge to the same limit at zero (and they do), then lim x-->0 f(x) is the same.</p>
<p>There's a pretty good discussion of the Squeeze Theorem on wikipedia at <a href="http://en.wikipedia.org/wiki/Squeeze_theorem%5B/url%5D">http://en.wikipedia.org/wiki/Squeeze_theorem</a> and the first example they use is similar to your problem.</p>