若函数f(x)={k+x2,x≤0x+1−1x,x>0f(x)=\begin{cases}k + x^2, & x \leq 0\\\dfrac{\sqrt{x + 1}-1}{x}, & x > 0\end{cases}f(x)=⎩⎨⎧k+x2,xx+1−1,x≤0x>0在(−∞,+∞)(-\infty,+\infty)(−∞,+∞)内连续，求kkk的值.
求极限limx→+∞x2+5xex\lim\limits_{x \to +\infty} \dfrac{x^2 + 5x}{e^{x}}x→+∞limexx2+5x.
f(x)=x2−3x+2f(x) = x^2 - 3x + 2f(x)=x2−3x+2在[−3,4][-3, 4][−3,4]上的最大值是（ ）.
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