Find the limit.
-$$\lim _ { \mathrm { P } \rightarrow ( 2,0 , - 2 ) } \mathrm { e } ^ { 2 } \sin ^ { - 1 } \left( - \frac { \mathrm { z } } { \mathrm { x } } \right)$$
A) $\frac { \pi } { 2 }$
B) 0
C) $\pi$
D) 1

Question

Quizplus · Accepted Answer

To solve this problem, we must use spherical coordinates. Converting Cartesian coordinates to spherical coordinates, we get:
\begin{align*}
r &= \sqrt{(x-2)^2 + y^2 + (z+2)^2} \
\theta &= \tan^{-1} \frac{z+2}{x-2} \
\phi &= \cos^{-1} \frac{y}{\sqrt{(x-2)^2 + y^2 + (z+2)^2}}
\end{align*}

As we approach $(2, 0, -2)$, we have $r \to 0$, $\theta \to \frac{3\pi}{4}$, and $\phi$ is constant at $\frac{\pi}{2}$. Therefore, in spherical coordinates, the limit becomes:
$$\lim_{r \to 0} e^2 \sin^{-1} \left(-\frac{\cos \phi}{\sin \theta}\right) = \lim_{r \to 0} e^2 \sin^{-1} \left(-\cos \phi \sqrt{\frac{x-2}{(x-2)^2 + (z+2)^2}}\right)$$

Since $\sin^{-1}(-x) = -\sin^{-1}(x)$, we can write the above limit as:
$$\lim_{r \to 0} -e^2 \sin^{-1} \left(\cos \phi \sqrt{\frac{x-2}{(x-2)^2 + (z+2)^2}}\right)$$

But note that as $\phi$ varies from $0$ to $2\pi$, the expression inside the square root doesn't change (since $y$ is fixed at $0$, and the denominator is just $r$). Therefore, we can treat $\phi$ as a constant and take the limit as $r \to 0$:
$$-e^2 \sin^{-1} \left(\cos \phi \sqrt{\lim_{r \to 0} \frac{x-2}{(x-2)^2 + (z+2)^2}}\right)$$

Now we must compute the remaining limit. In polar coordinates, we have $x = 2 + r \cos \theta$ and $z = -2 + r \sin \theta$, so:
\begin{align*}
\lim_{r \to 0} \frac{x-2}{(x-2)^2 + (z+2)^2} &= \lim_{r \to 0} \frac{r \cos \theta}{r^2 + 4r^2 \cos^2 \theta + 4r^2 \sin^2 \theta} \
&= \frac{\cos \theta}{5}
\end{align*}

Plugging this back into the earlier expression, we get:
$$-e^2 \sin^{-1} \left(\cos \phi \sqrt{\frac{\cos \theta}{5}}\right)$$

Finally, taking the limit as $\theta \to \frac{3\pi}{4}$ (and using the fact that $\cos(3\pi/4) = -\sin(\pi/4) = -\frac{1}{\sqrt{2}}$), we get:
$$-e^2 \sin^{-1} \left(\frac{\cos \phi}{\sqrt{10}}\right)$$

This limit does not depend on $r$ or $\theta$, so it is the same as the value of the function at the point $(2, 0, -2)$, which is $\boxed{\frac{\pi}{2}}$ (since $\cos \phi$ varies from $-1$ to $1$ as $\phi$ varies from $0$ to $2\pi$).

Find the Limit lim⁡P→(2,0,−2)e2sin⁡−1(−zx)\lim _ { \mathrm { P } \rightarrow ( 2,0 , - 2 ) } \mathrm { e } ^ { 2 } \sin ^ { - 1 } \left( - \frac { \mathrm { z } } { \mathrm { x } } \right)P→(2,0,−2)lim​e2sin−1(−xz​)

Find the Limit $\lim _ { \mathrm { P } \rightarrow ( 2,0 , - 2 ) } \mathrm { e } ^ { 2 } \sin ^ { - 1 } \left( - \frac { \mathrm { z } } { \mathrm { x } } \right)$