limx→0+1−cosx1+4x−1\lim\limits_{x\to0^{+}}\dfrac{1 - \cos\sqrt{x}}{\sqrt{1 + 4x}-1}x→0+lim1+4x−11−cosx.
limx→0∫0x2tantdtx4\lim\limits_{x\to0}\dfrac{\int_{0}^{x^{2}}\tan t\mathrm{d}t}{x^{4}}x→0limx4∫0x2tantdt.
{2x+a,x⩽0ln(1+2x),x>0\begin{cases}2x + a, & x\leqslant0\\\ln(1 + 2x), & x>0\end{cases}{2x+a,ln(1+2x),x⩽0x>0在点x=0x = 0x=0处连续,求aaa的值.
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