\[ \lim_{x \to 1} \dfrac{3}{x-1} = \dfrac{3}{1-1} = \frac{3}{0} \]

As $x \to 1^-$ from the left, the curve heads down toward $-\infty$. And as $x \to 1^+$ from the right, the curve heads up toward $\infty$. Hence this limit does not exist:
$\boxed{\text{DNE}}\quad \cmark$

We always first try Substitution:
$$\lim_{x \to 3} \frac{x-3}{x^2-9} = \frac{3-3}{9-9} = \frac{0}{0}$$
Because this limit is in the form of $\displaystyle{\frac{0}{0}}$, it is “indeterminate”—we don’t know what it actually equals. So we have more work to do.

The first thing you should always try when hitting such an indeterminate limit is to factor the numerator or denominator, and see if common term cancels. In this case we can factor the denominator:
\begin{align*}
\lim_{x \to 3} \frac{x-3}{x^2-9} &= \lim_{x \to 3}\frac{(x-3)}{(x-3)(x+3)} \\[12px]
&= \lim_{x \to 3}\frac{\cancel{(x-3)}}{\cancel{(x-3)}(x+3)} \\[12px]
&= \lim_{x \to 3} \frac{1}{x+3} \\[12px]
&= \frac{1}{3+3} = \frac{1}{6} \quad \cmark
\end{align*}
This function thus has the same graph as $y=\dfrac{1}{x+3}$, except that there is a hole at $x=3$ where our function is undefined. See the graph.

We first try substitution:
\[\lim_{x \to 1} \frac{x^2 + 4x -5}{x^2 – 1} = \frac{1 +4(1) – 5}{1-1} = \frac{0}{0}\]
Because this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is.
So let’s factor the numerator and the denominator:
\begin{align*}
\lim_{x \to 1} \frac{x^2 + 4x -5}{x^2 – 1} &= \lim_{x \to 1} \frac{(x+5)(x-1)}{(x-1)(x+1)}\\[8px]
&= \lim_{x \to 1} \frac{(x+5)\cancel{(x-1)}}{\cancel{(x-1)}(x+1)} \\[8px]
&= \lim_{x \to 1}\frac{x+5}{x+1} \\[8px]
&= \frac{1+5}{1+1} = \frac{6}{2} \\[8px]
&=3 \quad \cmark
\end{align*}

So let’s rationalize the denominator by using the usual approach of multiplying by its conjugate $ \dfrac{\sqrt{x} + 3}{\sqrt{x} + 3} = 1$ :
\begin{align*}
\lim_{x \to 9} \frac{x-9}{\sqrt{x}-3} &= \lim_{x \to 9} \left( \frac{x-9}{\sqrt{x}-3} \right) \left( \frac{\sqrt{x} + 3}{\sqrt{x} + 3} \right) \\[12px]
&= \lim_{x \to 9} \frac{(x-9)(\sqrt{x} + 3)}{(x – 9)} \\[12px]
&= \lim_{x \to 9} \frac{\cancel{(x-9)}(\sqrt{x} + 3)}{\cancel{(x – 9)}} \\[12px]
&= \lim_{x \to 9}\sqrt{x} + 3 \\[12px]
&= \sqrt{9} + 3 = 6 \quad \cmark
\end{align*}

We first try substitution:

\[ \lim_{x \to 3}\frac{\sqrt{x+1} -2}{x-3} = \frac{\sqrt{3+1} -2}{3-3} = \frac{\sqrt{4}-2}{3-3} = \frac{0}{0} \]

Since this limit is in the form $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is.  So let’s rationalize the expression:

\[ \begin{align*}
\lim_{x \to 3}\frac{\sqrt{x+1} -2}{x-3} &= \lim_{x \to 3}\frac{\sqrt{x+1} -2}{x-3} \cdot \frac{\sqrt{x+1} +2}{\sqrt{x+1} +2} \\[8px]
&= \lim_{x \to 3} \frac{\sqrt{x+1}\sqrt{x+1}+ 2\sqrt{x+1} – 2 \sqrt{x+1} -4}{(x-3)(\sqrt{x+1}+2)} \\[8px]
&= \lim_{x \to 3}\frac{(x+1)-4}{(x-3)(\sqrt{x+1}+2)}\\[8px]
&= \lim_{x \to 3}\frac{x-3}{(x-3)(\sqrt{x+1}+2)} \\[8px]
&= \lim_{x \to 3}\frac{1}{\sqrt{x+1}+2} \\[8px]
&= \frac{1}{\sqrt{3+1}+2} \\[8px]
&= \frac{1}{\sqrt{4}+2} = \frac{1}{2+2} \\[8px]
&= \frac{1}{4} \quad \cmark
\end{align*} \]

As always, we first try Substitution and simply plug $h=0$ into the expression:
$$\lim_{h \to 0}\dfrac{(h-5)^2 – 25}{h} = \dfrac{(0-5)^2 -25}{0} = \dfrac{25-25}{0} = \dfrac{0}{0}$$
Since the limit is in the form $\dfrac{0}{0},$ it is indeterminate—we don’t yet know what is it. So we have to do some work to turn the expression into a different form that’s more helpful.

Let’s try the simple move of expanding the quadratic in the numerator, and then see what happens:
\begin{align*}
\lim_{h \to 0}\dfrac{(h-5)^2 – 25}{h} &= \lim_{h \to 0}\dfrac{(h^2 -10h + 25) – 25}{h} \\[12px]
&= \lim_{h \to 0}\dfrac{h^2 – 10h}{h} \\[12px]
&= \lim_{h \to 0}\dfrac{h(h – 10)}{h} \\[12px]
&= \lim_{h \to 0}\dfrac{\cancel{h}(h-10)}{\cancel{h}} \\[12px]
&= \lim_{h \to 0}(h – 10) \\[12px]
&= 0 – 10 = -10 \quad \cmark
\end{align*}
Notice that the problematic h in the denominator cancelled out, just as we were hoping would happen. We can then use simple Substitution to finish.

As always, we first try Substitution:
$$\lim_{x \to 0}\dfrac{1- \cos(x)}{\sin^2(x)} = \frac{1- \cos(0)}{\sin^2(0)} = \frac{1-1}{0} = \frac{0}{0}$$
Yet again we obtain $\dfrac{0}{0}$, that indeterminate result: we don’t yet know what the limit is. We have more work to do.

Seeing the $\sin^2(x)$ there in the denominator makes us think about the trig identity above, which we can rewrite as
$$ \sin^2(x) = 1 – \cos^2(x) $$
Hence

\begin{align*} \lim_{x \to 0}\frac{1- \cos(x)}{\sin^2(x)} &= \lim_{x \to 0}\frac{1- \cos(x)}{1 – \cos^2(x)} \\[8px] \text{We can factor the denominator, } &\text{so let’s do it} \\ \text{and hope that something cancels:} \\[8px] &= \lim_{x \to 0}\frac{1- \cos(x)}{(1 – \cos(x))(1 + \cos(x))} \\[8px] \text{Yep, we can cancel some terms!} \\[8px] &= \lim_{x \to 0}\frac{\cancel{1- \cos(x)}}{\cancel{(1 – \cos(x))}(1 + \cos(x))} \\[8px] &= \lim_{x \to 0} \frac{1}{1 + \cos(x)} \\[8px] \text{And now Substitution } \text{to finish:} \\[8px] &= \frac{1}{1 + \cos(0)} \\[8px] &= \frac{1}{1+1} = \frac{1}{2} \quad \cmark \end{align*}

The expression in the question reminds us of the first “Special Limit,”
$$\lim_{x \to 0}\frac{\sin(x)}{x} = 1$$
But it isn’t quite the same, because in our expression the argument of sin that’s in the numerator (5x) doesn’t match what’s in the denominator (x). That is, since we have $\sin(5x)$ in the numerator, we need $5x$ in the denominator.

So let’s multiply the expression by $\dfrac{5}{5}$, and then do some rearranging:
\begin{align*}
\lim_{x \to 0}\dfrac{\sin(5x)}{x} &= \lim_{x \to 0} \frac{5}{5} \cdot \dfrac{\sin(5x)}{x} \\[8px]
&= 5 \cdot \lim_{x \to 0} \frac{\sin(5x)}{5x} &\text{[Recall $\displaystyle{\lim_{x \to 0}\dfrac{\sin(\text{whatever})}{(\text{the same whatever})} =1 }$]}\\[8px]
&= 5 \cdot 1 = 5 \quad \cmark
\end{align*}