Wednesday, 18 June 2014

calculus - One sided limit with ex and sin(x)



Assuming that exea and sin(x)sin(a)



evaluate the limit as lim



To me, based on the first sentence. I could just plug in a and get my limit which does not exist at the point x=0 However the back of the book says infinity so I am at a loss as to how to prove this.



This is a real analysis class but this is isn't one of the problems requiring us to prove with definitions but it would be helpful to learn the algebra and the theory. I don't understand as to how I can apply the definition of a function converging to infinity to prove this. Plus I don't see how that could help? How am I supposed to know it goes to infinity


Answer




When x is close enough to 0, then $-1.1 so that
\exp(-1.1)<\exp(x^2+2x-1)



and hence the function in your limit is bounded from below by \frac{\exp(-1.1)}{\sin x} which blows up at x=0 since \sin0=0.


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