QuarkCharmer said:Is that sin(2x^2), sin^2(2x), sin(2)x^2 ?
Dick said:Or is it 3sin(2x^2-6x^2+4x^6)? The question isn't very grammatical. And however I rearrange it I still can't figure out how to pull 4/5 as a limit out.
vela said:It helps to use the substitution u=x2 to get
[tex]\lim_{u \to 0}\frac{3 \sin 2u - 6u + 4u^3}{u^5}[/tex]Then three applications of the Hospital rule gets you to a limit you can evaluate.
When we say that x is approaching 0, it means that the value of x is getting closer and closer to 0, but is not actually equal to 0.
The limit as x approaches 0 is important because it helps us understand the behavior of a function near the point x=0. It can also help us determine the continuity and differentiability of a function at x=0.
To find the limit as x approaches 0, we substitute 0 into the function and simplify. If the resulting value is undefined or infinite, we use algebraic techniques or L'Hopital's rule to evaluate the limit.
A one-sided limit only considers the values of the function as x approaches 0 from one direction (either the left or the right). A two-sided limit takes into account the values of the function as x approaches 0 from both the left and the right.
Yes, it is possible for the limit as x approaches 0 to exist even if the function is not defined at x=0. This is because the limit is concerned with the behavior of the function near x=0, not necessarily at x=0.