(a) lim ( x , y ) → ( 2 , − 1 ) ( x y + x 2 ) = 2 ( − 1 ) + 2 2 = 2 \displaystyle \lim_{(x,y)\rightarrow(2,-1)}\ \big(xy+x^2\big)=2(-1)+2^2=2 ( x , y ) → ( 2 , − 1 ) lim ( x y + x 2 ) = 2 ( − 1 ) + 2 2 = 2
(b) Switching to polar coordinates,
lim ( x , y ) → ( 0 , 0 ) x x 2 + y 2 = lim r → 0 + 0 ≤ θ < 2 π r cos θ r 2 = lim r → 0 + 0 ≤ θ < 2 π cos θ r \begin{aligned}
\lim_{(x,y)\rightarrow(0,0)}\ \frac{x}{x^2+y^2}
&=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{r\cos\theta}{r^2}
=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{\cos\theta}{r}
\end{aligned} ( x , y ) → ( 0 , 0 ) lim x 2 + y 2 x = 0 ≤ θ < 2 π r → 0 + lim r 2 r cos θ = 0 ≤ θ < 2 π r → 0 + lim r cos θ
which does not exist, since, for example,
lim r → 0 + θ = 0 cos θ r = lim r → 0 + 1 r = + ∞ \lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{\theta=0}}\ \frac{\cos\theta}{r}
=\lim_{r\rightarrow0^+}\ \frac{1}{r}
=+\infty θ = 0 r → 0 + lim r cos θ = r → 0 + lim r 1 = + ∞
while if θ = π \theta=\pi θ = π
lim r → 0 + θ = π cos θ r = lim r → 0 + − 1 r = − ∞ \lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{\theta=\pi}}\ \frac{\cos\theta}{r}
=\lim_{r\rightarrow0^+}\ \frac{-1}{r}
=-\infty θ = π r → 0 + lim r cos θ = r → 0 + lim r − 1 = − ∞
(c) Switching to polar coordinates,
lim ( x , y ) → ( 0 , 0 ) x 2 x 2 + y 2 = lim r → 0 + 0 ≤ θ < 2 π r 2 cos 2 θ r 2 = lim r → 0 + 0 ≤ θ < 2 π cos 2 θ \begin{aligned}
\lim_{(x,y)\rightarrow(0,0)}\ \frac{x^2}{x^2+y^2}
&=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{r^2\cos^2\theta}{r^2}
=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \cos^2\theta
\end{aligned} ( x , y ) → ( 0 , 0 ) lim x 2 + y 2 x 2 = 0 ≤ θ < 2 π r → 0 + lim r 2 r 2 cos 2 θ = 0 ≤ θ < 2 π r → 0 + lim cos 2 θ
which does not exist, since, for example,
lim r → 0 + θ = 0 cos 2 θ = lim r → 0 + 1 = 1 \lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{\theta=0}}\ \cos^2\theta
=\lim_{r\rightarrow0^+}\ 1
= 1 θ = 0 r → 0 + lim cos 2 θ = r → 0 + lim 1 = 1
while if θ = π 2 \theta=\frac{\pi}{2} θ = 2 π
lim r → 0 + θ = π / 2 cos 2 θ = lim r → 0 + 0 = 0 \lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{\theta=\pi/2}}\ \cos^2\theta
=\lim_{r\rightarrow0^+}\ 0
=0 θ = π /2 r → 0 + lim cos 2 θ = r → 0 + lim 0 = 0
(d) Switching to polar coordinates,
lim ( x , y ) → ( 0 , 0 ) x 3 x 2 + y 2 = lim r → 0 + 0 ≤ θ < 2 π r 3 cos 3 θ r 2 = lim r → 0 + 0 ≤ θ < 2 π r cos 3 θ = 0 \begin{aligned}
\lim_{(x,y)\rightarrow(0,0)}\ \frac{x^3}{x^2+y^2}
&=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{r^3\cos^3\theta}{r^2}
=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ r\cos^3\theta
=0
\end{aligned} ( x , y ) → ( 0 , 0 ) lim x 2 + y 2 x 3 = 0 ≤ θ < 2 π r → 0 + lim r 2 r 3 cos 3 θ = 0 ≤ θ < 2 π r → 0 + lim r cos 3 θ = 0
since ∣ cos θ ∣ ≤ 1 |\cos\theta|\le 1 ∣ cos θ ∣ ≤ 1 θ \theta θ
(e) Switching to polar coordinates,
lim ( x , y ) → ( 0 , 0 ) x 2 y 2 x 2 + y 4 = lim r → 0 + 0 ≤ θ < 2 π r 2 cos 2 θ r 2 sin 2 θ r 2 cos 2 θ + r 4 sin 4 θ = lim r → 0 + 0 ≤ θ < 2 π r 2 sin 2 θ cos 2 θ cos 2 θ + r 2 sin 4 θ = 0 \begin{aligned}
\lim_{(x,y)\rightarrow(0,0)}\ \frac{x^2y^2}{x^2+y^4}
&=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\
\frac{r^2\cos^2\theta\ r^2\sin^2\theta}
{r^2\cos^2\theta+r^4\sin^4\theta}
=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ r^2\sin^2\theta
\frac{\cos^2\theta}{\cos^2\theta+r^2\sin^4\theta} \\
&= 0
\end{aligned} ( x , y ) → ( 0 , 0 ) lim x 2 + y 4 x 2 y 2 = 0 ≤ θ < 2 π r → 0 + lim r 2 cos 2 θ + r 4 sin 4 θ r 2 cos 2 θ r 2 sin 2 θ = 0 ≤ θ < 2 π r → 0 + lim r 2 sin 2 θ cos 2 θ + r 2 sin 4 θ cos 2 θ = 0
Here, we used that
∣ sin 2 θ cos 2 θ cos 2 θ + r 2 sin 4 θ ∣ ≤ cos 2 θ cos 2 θ + r 2 sin 4 θ ≤ { cos 2 θ cos 2 θ if cos θ ≠ 0 0 if cos θ = 0 } ≤ 1 \left|\sin^2\theta\frac{\cos^2\theta} {\cos^2\theta+r^2\sin^4\theta}\right|
\le \frac{\cos^2\theta} {\cos^2\theta+r^2\sin^4\theta}
\le \left.\begin{cases}
\frac{\cos^2\theta} {\cos^2\theta}&\text{ if }\cos\theta\ne 0 \\
0 &\text { if }\cos\theta =0
\end{cases}
\right\}
\le 1 sin 2 θ cos 2 θ + r 2 sin 4 θ cos 2 θ ≤ cos 2 θ + r 2 sin 4 θ cos 2 θ ≤ { c o s 2 θ c o s 2 θ 0 if cos θ = 0 if cos θ = 0 } ≤ 1
for all r > 0 r>0 r > 0
(f) To start, observe that
lim ( x , y ) → ( 0 , 0 ) ( sin x ) ( e y − 1 ) x y = [ lim x → 0 sin x x ] [ lim y → 0 e y − 1 y ] \begin{aligned}
\lim_{(x,y)\rightarrow (0,0)}\
\frac{(\sin x)\left(e^y-1\right)}{xy}
=\left[\lim_{x\rightarrow 0}\
\frac{\sin x}{x}\right]
\left[\lim_{y\rightarrow 0}\
\frac{e^y-1}{y}\right]
\end{aligned} ( x , y ) → ( 0 , 0 ) lim x y ( sin x ) ( e y − 1 ) = [ x → 0 lim x sin x ] [ y → 0 lim y e y − 1 ]
We may evaluate [ lim x → 0 sin x x ] \displaystyle \left[\lim_{x\rightarrow 0}\
\frac{\sin x}{x}\right] [ x → 0 lim x sin x ]
lim x → 0 sin x x = lim x → 0 sin x − sin 0 x − 0 = d d x sin x ∣ x = 0 = cos x ∣ x = 0 = 1 \lim_{x\rightarrow 0}\ \frac{\sin x}{x}
=\lim_{x\rightarrow 0}\ \frac{\sin x-\sin 0}{x-0}
=\frac{\mathrm{d}}{\mathrm{d}x}\sin x\bigg|_{x=0}
=\cos x\Big|_{x=0}=1 x → 0 lim x sin x = x → 0 lim x − 0 sin x − sin 0 = d x d sin x x = 0 = cos x x = 0 = 1
Similarly, we may evaluate [ lim y → 0 e y − 1 y ] \displaystyle \left[\lim_{y\rightarrow 0}\
\frac{e^y-1}{y}\right] [ y → 0 lim y e y − 1 ]
lim y → 0 e y − 1 y = lim y → 0 e y − e 0 y − 0 = d d y e y ∣ y = 0 = e y ∣ y = 0 = 1 \lim_{y\rightarrow 0}\ \frac{e^y-1}{y}
=\lim_{y\rightarrow 0}\ \frac{e^y-e^0}{y-0}
=\frac{\mathrm{d}}{\mathrm{d}y}e^y\bigg|_{y=0}
=e^y\Big|_{y=0}=1 y → 0 lim y e y − 1 = y → 0 lim y − 0 e y − e 0 = d y d e y y = 0 = e y y = 0 = 1
So all together
lim ( x , y ) → ( 0 , 0 ) ( sin x ) ( e y − 1 ) x y = [ lim x → 0 sin x x ] [ lim y → 0 e y − 1 y ] = [ 1 ] [ 1 ] = 1 \begin{aligned}
\lim_{(x,y)\rightarrow (0,0)}\
\frac{(\sin x)\left(e^y-1\right)}{xy}
=\left[\lim_{x\rightarrow 0}\
\frac{\sin x}{x}\right]
\left[\lim_{y\rightarrow 0}\
\frac{e^y-1}{y}\right]
=[1]\ [1]=1
\end{aligned} ( x , y ) → ( 0 , 0 ) lim x y ( sin x ) ( e y − 1 ) = [ x → 0 lim x sin x ] [ y → 0 lim y e y − 1 ] = [ 1 ] [ 1 ] = 1