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Webกากสุดในพรอน4 @tanxlive WebJun 12, 2024 · lim x→0 sinx x = 1. Now, let's look at our problem and manipulate it a bit: lim x→0 tanx x. = lim x→0 sinx/cosx x. = lim x→0 (sinx x) cosx. = lim x→0 ( sinx x) ⋅ ( 1 cosx) Remember that the limit of a product is the product of the limits, if both limits are defined. = ( lim x→0 sinx x) ⋅ ( lim x→0 1 cosx) = 1 ⋅ 1 cos0. WebWhat follows is one way to proceed, assuming you take log to refer to the natural logarithm. Recall that ∫ log(u) du = ulog(u) - u + C, where C is any real number. Using the substitution u = x + 1, du = dx, we may write ∫ log(x + 1) dx = ∫ log(u) du = ulog(u) - u + C.Now we may substitute u = x + 1 back into the last expression to arrive at the answer: oxy of pa inc