You don't need tricks to solve the limit of (x^2)sin(2x)/sqrt(x) as x->0, only to know equivalent infinitesimals, so that sin(2x) can be replaced by 2x in this

You don't need tricks to solve the limit of (x^2)sin(2x)/sqrt(x) as x->0, only to know equivalent infinitesimals, so that sin(2x) can be replaced by 2x in this limit and 2x^3/sqrt(x) can be simplified to 2x^(3-1/2) so the limit is zero. Moreover, you should know that x^2 and sin(2x) are really smaller compared to sqrt(x) so that the limit is zero. Knowing how to compare two infinitesimals can help you to replace a difficult process for an easier one, just like in differential equations sin(x) is replaced by x for small vibrations. That kind of knowledge allows you obtain an approximate solution in many real processes. So I don't consider that knowledge a bag of tricks. Furthermore, replacing sin(2x) for 2x is just the process of taking the linear approximation (the tangent line near x=0 for y=sin(2x)), so this is just an example of the power of calculus to replace a difficult question with an easier one. Also you learn to explore the world near a point (like a miscroscope) by the process of taking the linear approximation or taylor expansion of a function near a point, then you iterate this procedure to nearby points just like what happens when you solve a differential equation by Euler method. So I should say that there is a framework to solve problems via approximate solutions and computing limits is just a good first step to enter the realm of Calculus.