$\displaystyle \lim_{x\rightarrow 1} h(x)$ exists if and only if $\displaystyle \lim_{x\rightarrow 1^{-}} h(x) = \lim_{x\rightarrow 1^{+}} h(x)$. As can be seen from the diagram, $\displaystyle \lim_{x\rightarrow 1^{-}} h(x) = 1$, but $\displaystyle \lim_{x\rightarrow 1^{+}} h(x) = -1$. Since $\displaystyle \lim_{x\rightarrow 1^{-}} h(x) \neq \lim_{x\rightarrow 1^{+}} h(x)$,   $\displaystyle \lim_{x\rightarrow 1} h(x)$ does not exist.

$\displaystyle \lim_{x\rightarrow 1} f(x)$ exists if and only if $\displaystyle \lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)$;

the actual value of $\displaystyle f(1)$ is irrelevant, as is whether $\displaystyle f$ is continuous there.

As can be seen,

$\displaystyle \lim_{x\rightarrow 1^{+}} f(x) = 1$ and $\displaystyle \lim_{x\rightarrow 1^{-}} f(x) = 1$;

therefore, $\displaystyle \lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)$,

and $\displaystyle \lim_{x\rightarrow 1} f(x)$ exists.

The first step to determine continuity at a point is to determine if the function is defined at that point. When we substitute in 3 for $\displaystyle x$, we get 18 as our $\displaystyle y$-value. $\displaystyle x=3$ is thus defined for this function.

The next step is determine if the limit of the function is defined at that point. This means that the left-hand limit must be equal to the right-hand limit at $\displaystyle x=3$. Substitution reveals the following:

$\displaystyle \lim_{x\rightarrow 3^{-}}f(x)=(3)^{2}+4(3)-3=9+12-3=18$

$\displaystyle \lim_{x\rightarrow 3^{+}}f(x)=10(3)-12=18$

Both sides of the function, therefore, approach a $\displaystyle y$-value of 18.

Finally, we must ensure that the curve is smooth by checking the limit of the derivative of both sides.

$\displaystyle \lim_{x\rightarrow 3^{-}}f^{'}(x)=2(3)+4=10$

$\displaystyle \lim_{x\rightarrow 3^{+}}f^{'}(x)=10$

Since the function passes all three tests, it is continuous.

$\displaystyle \lim_{x\rightarrow 1}g(x)$ exists if and only if $\displaystyle \lim_{x\rightarrow 1^{+}} g(x) = \lim_{x\rightarrow 1^{-}} g(x)$; the actual value of $\displaystyle f(1)$ is irrelevant.

As can be seen, $\displaystyle \lim_{x\rightarrow 1^{+}} g(x) = 1$ and $\displaystyle \lim_{x\rightarrow 1^{-}}g(x) = 1$; therefore, $\displaystyle \lim_{x\rightarrow 1^{+}} g(x) = \lim_{x\rightarrow 1^{-}} g(x)$, and $\displaystyle \lim_{x\rightarrow 1} g(x)$  exists.

Given the polar coordinates $\displaystyle (r,\theta )$, the  $\displaystyle y$-coordinate is  $\displaystyle y= r \sin \theta$.  We can find this coordinate by substituting $\displaystyle r = 2.1, \theta = \frac{7\pi }{3}$:

$\displaystyle y = r \sin \theta = 2.1 \cdot \sin \frac{7\pi }{3} \approx 2.1 \cdot 0.8660 \approx 1.82$

Given the polar coordinates $\displaystyle (r,\theta )$, the  $\displaystyle x$-coordinate is  $\displaystyle x= r \cos \theta$. We can find this coordinate by substituting $\displaystyle r = 2.1, \theta = \frac{7\pi }{3}$:

$\displaystyle x = r \cos \theta = 2.1 \cdot \cos \frac{7\pi }{3} = 2.1 \cdot 0.5 = 1.05$

Given the polar coordinates $\displaystyle (r,\theta )$, the  $\displaystyle y$-coordinate is  $\displaystyle y= r \sin \theta$. We can find this coordinate by substituting $\displaystyle r = 1.2, \theta = \frac{2\pi }{5}$:

$\displaystyle y = r \cos \theta = 1.2 \cdot \sin \frac{2\pi }{5} \approx 1.2 \cdot 0.9511 \approx 1.14$

Given the polar coordinates $\displaystyle (r,\theta )$, the  $\displaystyle x$-coordinate is  $\displaystyle x= r \cos \theta$. We can find this coordinate by substituting $\displaystyle r = 1.2, \theta = \frac{2\pi }{5}$:

$\displaystyle x = r \cos \theta = 1.2 \cdot \cos \frac{2\pi }{5} \approx 1.2 \cdot 0.3090 \approx 0.37$

We need to subtract the $\displaystyle x$-coordinate and the $\displaystyle y$-coordinates to solve for a vector when given its initial and terminal coordinates:

Initial pt: $\displaystyle (-1,4)$

Terminal pt: $\displaystyle (2,2)$

Vector: $\displaystyle < 2-(-1),2-4>$

Vector: $\displaystyle < 3,-2>$