夹逼准则
例一：求极限limx→0y→0sin(x2y+y4)x2+y2\lim \limits_{x \rightarrow 0 \atop y \rightarrow 0} \frac{\sin \left(x^{2} y+y^{4}\right)}{x^{2}+y^{2}}y→0x→0limx2+y2sin(x2y+y4)
解析：因为∣sinx∣≤∣x∣|\sin x| \leq|x|∣sinx∣≤∣x∣，因为有
0≤∣sin(x2y+y4)x2+y2∣≤∣x2y+y4x2+y2∣0 \leq\left|\frac{\sin \left(x^{2} y+y^{4}\right)}{x^{2}+y^{2}}\right| \leq\left|\frac{x^{2} y+y^{4}}{x^{2}+y^{2}}\right| 0≤∣∣∣∣∣x2+y2sin(x2y+y4)∣∣∣∣∣≤∣∣∣∣x2+y2x2y+y4∣∣∣∣
又因为
∣x2y+y4x2+y2∣≤x2x2+y2×∣y∣+y2x2+y2×y2≤∣y∣+y2→0\begin{aligned} \left|\frac{x^{2} y+y^{4}}{x^{2}+y^{2}}\right| & \leq \frac{x^{2}}{x^{2}+y^{2}} \times|y|+\frac{y^{2}}{x^{2}+y^{2}} \times y^{2} \\ & \leq|y|+y^{2} \rightarrow 0 \end{aligned} ∣∣∣∣x2+y2x2y+y4∣∣∣∣≤x2+y2x2×∣y∣+x2+y2y2×y2≤∣y∣+y2→0
由夹逼准则知，极限为0
例2：求极限limx→+∞y→+∞(xyx2+y2)x2\lim \limits_{x \rightarrow+\infty \atop y \rightarrow+\infty}\left(\frac{x y}{x^{2}+y^{2}}\right)^{x^{2}}y→+∞x→+∞lim(x2+y2xy)x2
解析：注意到
0≤xyx2+y2≤12(x2+y2)x2+y2=120 \leq \frac{x y}{x^{2}+y^{2}} \leq \frac{\frac{1}{2}\left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}=\frac{1}{2} 0≤x2+y2xy≤x2+y221(x2+y2)=21
所以
0≤(xyx2+y2)x2≤(12)x2→00 \leq\left(\frac{x y}{x^{2}+y^{2}}\right)^{x^{2}} \leq\left(\frac{1}{2}\right)^{x^{2}} \rightarrow 0 0≤(x2+y2xy)x2≤(21)x2→0
由夹逼准则知极限为0
例三求极限limx→∞y→∞x+yx2−xy+y2\lim \limits_{x \rightarrow \infty \atop y \rightarrow \infty} \frac{x+y}{x^{2}-x y+y^{2}}y→∞x→∞limx2−xy+y2x+y
解法一：由于
∣x+yx2−xy+y2∣≤∣1y+1x∣∣xy−1+yx∣≤∣1y+1x∣∣xy+yx∣−1≤∣1y+1x∣→0\left|\frac{x+y}{x^{2}-x y+y^{2}}\right| \leq \frac{\left|\frac{1}{y}+\frac{1}{x}\right|}{\left|\frac{x}{y}-1+\frac{y}{x}\right|} \leq \frac{\left|\frac{1}{y}+\frac{1}{x}\right|}{\left|\frac{x}{y}+\frac{y}{x}\right|-1} \leq\left|\frac{1}{y}+\frac{1}{x}\right| \rightarrow 0 ∣∣∣∣x2−xy+y2x+y∣∣∣∣≤∣∣∣yx−1+xy∣∣∣∣∣∣y1+x1∣∣∣≤∣∣∣yx+xy∣∣∣−1∣∣∣y1+x1∣∣∣≤∣∣∣∣y1+x1∣∣∣∣→0
解法二：由于
∣x+yx2−xy+y2∣≤2∣x+y∣x2+y2≤2∣x∣+∣y∣x2+y2≤2(1∣x∣+1∣y∣)→0\left|\frac{x+y}{x^{2}-x y+y^{2}}\right| \leq \frac{2|x+y|}{x^{2}+y^{2}} \leq 2 \frac{|x|+|y|}{x^{2}+y^{2}} \leq 2\left(\frac{1}{|x|}+\frac{1}{|y|}\right) \rightarrow 0 ∣∣∣∣x2−xy+y2x+y∣∣∣∣≤x2+y22∣x+y∣≤2x2+y2∣x∣+∣y∣≤2(∣x∣1+∣y∣1)→0
故由夹逼准则知极限为0
解法三：
注意到x2+y2−xy≥2xy−xy=xyx^{2}+y^{2}-x y \geq 2 x y-x y=x yx2+y2−xy≥2xy−xy=xy
由于
∣x+yx2−xy+y2∣≤∣x+yxy∣≤(1∣y∣+1∣x∣)→0\left|\frac{x+y}{x^{2}-x y+y^{2}}\right| \leq\left|\frac{x+y}{x y}\right| \leq\left(\frac{1}{|y|}+\frac{1}{|x|}\right) \rightarrow 0 ∣∣∣∣x2−xy+y2x+y∣∣∣∣≤∣∣∣∣xyx+y∣∣∣∣≤(∣y∣1+∣x∣1)→0
所以极限为0
极坐标
例题一：求极限lim(x,y)→(0,0)x3+y3x2+y2\lim \limits_{(x, y) \rightarrow(0,0)} \frac{x^{3}+y^{3}}{x^{2}+y^{2}}(x,y)→(0,0)limx2+y2x3+y3
解析：
令x=ρcosθ,y=ρsinθx=\rho \cos \theta, y=\rho \sin \thetax=ρcosθ,y=ρsinθ，则
lim(x,y)→(0,0)x3+y3x2+y2=limρ→0ρ3(cos3θ+sin3θ)ρ2=limρ→0ρ(cos3θ+sin3θ)=0\lim _{(x, y) \rightarrow(0,0)} \frac{x^{3}+y^{3}}{x^{2}+y^{2}}=\lim _{\rho \rightarrow 0} \frac{\rho^{3}\left(\cos ^{3} \theta+\sin ^{3} \theta\right)}{\rho^{2}}=\lim _{\rho \rightarrow 0} \rho\left(\cos ^{3} \theta+\sin ^{3} \theta\right)=0 (x,y)→(0,0)limx2+y2x3+y3=ρ→0limρ2ρ3(cos3θ+sin3θ)=ρ→0limρ(cos3θ+sin3θ)=0
化为一元函数
例题一：求极限limx→+∞y→+∞(x2+y2)e−(x+y)\lim \limits_{x \rightarrow+\infty \atop y \rightarrow+\infty}\left(x^{2}+y^{2}\right) e^{-(x+y)}y→+∞x→+∞lim(x2+y2)e−(x+y)
由于0<(x2+y2)ex+y=x2ex+y+y2ex+y≤x2ex+y2ey0<\frac{\left(x^{2}+y^{2}\right)}{e^{x+y}}=\frac{x^{2}}{e^{x+y}}+\frac{y^{2}}{e^{x+y}} \leq \frac{x^{2}}{e^{x}}+\frac{y^{2}}{e^{y}}0<ex+y(x2+y2)=ex+yx2+ex+yy2≤exx2+eyy2
易知x2ex、y2ey\frac{x^{2}}{e^{x}}、\frac{y^{2}}{e^{y}}exx2、eyy2都为0，所以极限为0
例题二：求极限lim(x,y)→(0,0)x2ln(x2+y2)=0\lim \limits_{(x, y) \rightarrow(0,0)} x^{2} \ln \left(x^{2}+y^{2}\right)=0(x,y)→(0,0)limx2ln(x2+y2)=0
解析：因为
lim(x,y)→(0,0)x2ln(x2+y2)=lim(x,y)→(0,0)x2x2+y2(x2+y2)ln(x2+y2)\lim \limits_{(x, y) \rightarrow(0,0)} x^{2} \ln \left(x^{2}+y^{2}\right)=\lim \limits_{(x, y) \rightarrow(0,0)} \frac{x^{2}}{x^{2}+y^{2}}\left(x^{2}+y^{2}\right) \ln \left(x^{2}+y^{2}\right)(x,y)→(0,0)limx2ln(x2+y2)=(x,y)→(0,0)limx2+y2x2(x2+y2)ln(x2+y2)
令x2+y2=t\sqrt{x^{2}+y^{2}}=tx2+y2=t
则有
lim(x,y)→(0,0)(x2+y2)ln(x2+y2)=limt→0+tlnt=limt→0+lnt1/t=limt→0+1/t−1/t2=0\begin{aligned} \lim _{(x, y) \rightarrow(0,0)}\left(x^{2}+y^{2}\right) \ln \left(x^{2}+y^{2}\right) &=\lim _{t \rightarrow 0^{+}} t \ln t \\ &=\lim _{t \rightarrow 0^{+}} \frac{\ln t}{1 / t}=\lim _{t \rightarrow 0^{+}} \frac{1 / t}{-1 / t^{2}}=0 \end{aligned} (x,y)→(0,0)lim(x2+y2)ln(x2+y2)=t→0+limtlnt=t→0+lim1/tlnt=t→0+lim−1/t21/t=0
例题三：求极限lim(x,y)→(0,0)xln(x2+y2)\lim \limits_{(x, y) \rightarrow(0,0)} x \ln \left(x^{2}+y^{2}\right)(x,y)→(0,0)limxln(x2+y2)
解析：因为
limx→0y→0xln(x2+y2)=2limx→0y→0xx2+y2x2+y2lnx2+y2\lim \limits_{x \rightarrow 0 \atop y \rightarrow 0} x \ln \left(x^{2}+y^{2}\right)=2 \lim \limits_{x \rightarrow 0 \atop y \rightarrow 0} \frac{x}{\sqrt{x^{2}+y^{2}}} \sqrt{x^{2}+y^{2}} \ln \sqrt{x^{2}+y^{2}}y→0x→0limxln(x2+y2)=2y→0x→0limx2+y2xx2+y2lnx2+y2
令x2+y2=t\sqrt{x^{2}+y^{2}}=tx2+y2=t，那么
lim(x,y)→(0,0)x2+y2lnx2+y2=limt→0+tlnt=limt→0+lnt1/t=limt→0+1/t−1/t2=0\begin{aligned} \lim _{(x, y) \rightarrow(0,0)} \sqrt{x^{2}+y^{2}} \ln \sqrt{x^{2}+y^{2}} &=\lim _{t \rightarrow 0^{+}} t \ln t \\ &=\lim _{t \rightarrow 0^{+}} \frac{\ln t}{1 / t}=\lim _{t \rightarrow 0^{+}} \frac{1 / t}{-1 / t^{2}}=0 \end{aligned} (x,y)→(0,0)limx2+y2lnx2+y2=t→0+limtlnt=t→0+lim1/tlnt=t→0+lim−1/t21/t=0
所以lim(x,y)→(0,0)xln(x2+y2)=0\lim \limits_{(x, y) \rightarrow(0,0)} x \ln \left(x^{2}+y^{2}\right)=0(x,y)→(0,0)limxln(x2+y2)=0
例题四：求极限lim(x,y)→(0,0)x2+y2−sinx2+y2(x2+y2)3/2\lim \limits_{(x, y) \rightarrow(0,0)} \frac{\sqrt{x^{2}+y^{2}}-\sin \sqrt{x^{2}+y^{2}}}{\left(x^{2}+y^{2}\right)^{3 / 2}}(x,y)→(0,0)lim(x2+y2)3/2x2+y2−sinx2+y2
解析：
lim(x,y)→(0,0)x2+y2−sinx2+y2(x2+y2)3/2x2+y2=ρρ→0ρ−sinρρ3=limρ→0ρ−(ρ−16ρ3+o(ρ3))ρ3=16\begin{aligned} \lim _{(x, y) \rightarrow(0,0)} \frac{\sqrt{x^{2}+y^{2}}-\sin \sqrt{x^{2}+y^{2}}}{\left(x^{2}+y^{2}\right)^{3 / 2}} & \frac{\sqrt{x^{2}+y^{2}}=\rho}{\rho \rightarrow 0} \frac{\rho-\sin \rho}{\rho^{3}} \\ &=\lim _{\rho \rightarrow 0} \frac{\rho-\left(\rho-\frac{1}{6} \rho^{3}+o\left(\rho^{3}\right)\right)}{\rho^{3}}=\frac{1}{6} \end{aligned} (x,y)→(0,0)lim(x2+y2)3/2x2+y2−sinx2+y2ρ→0x2+y2=ρρ3ρ−sinρ=ρ→0limρ3ρ−(ρ−61ρ3+o(ρ3))=61
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