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How does the SiNx power e of limx approaching 0 x come out I know the zero power of the answer e, but e doesn't know how to get it out
lim(x→0)x^sinx
=lim(x→0)e^(sinxlnx)
=lim(x→0)e^(xlnx)
=lim(x→0)e^(lnx/x^-1)
=lim(x→0)e^(-1/x/x^(-2))
=lim(x→0)e^(-x)
=1
Limx - > the limit of x power of infinite 4?
Also: the limit of 1 / x power of 2 at x = 0? The answer is that there is a left limit. But why?
Because the left limit at x = 0 is that x is infinitely close to 0 from the left side of the coordinate axis. When x is infinitely close to 0, you should think about how much the value of X should be now, and it should be negative [infinitesimal] (note that the infinitesimal here is for adding an absolute value to x, and so is the infinity behind. You can draw a sit
When x approaches zero, what is the limit of Ln (1 + 2x) divided by X
lim{x->0}ln(1+2x)/x=lim{x->0}2x/x=2.
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