X tends to 0, the limit of sin1 / x ^ 2-1 / sin2x (not SiNx ^ 2, sin2x)

X tends to 0, the limit of sin1 / x ^ 2-1 / sin2x (not SiNx ^ 2, sin2x)


Whether it is (SiNx) ^ 2 or sin (2x), the limit is ∞. Because 1 / sin2x is ∞ when x tends to 0, and sin1 / x ^ 2 is a bounded function, the difference must be ∞



When x tends to 0, why can't sin1 / X be equal to 1 / X in the limit of (square times sin1 / x) divided by SiNx?


Because when x tends to zero, SiNx can be equivalent to X
Here, 1 / X goes to infinity, so it doesn't work



Does the convergence function have bound and limit? Then does 1 / X converge? Why can X - > infinity get 0
The concept is vague, don't copy


The convergent sequence must be bounded
Read carefully, the book never said that the convergence function is bounded, only that the convergence function is locally bounded, the so-called locally bounded can be simply understood as bounded near the convergence point, away from it may be unbounded