You always want to try direct substitution first.
\(\displaystyle{\lim_{x \rightarrow 3} \left[ \frac{5x^2-8x-13}{x^2-5} \right] = }\) \(\displaystyle{ \frac{5(9)-8(3)-13}{9-5} = }\) \(\displaystyle{ \frac{45-24-13}{4} = }\) \(\displaystyle{ \frac{8}{4} = 2 }\)
In this case, the result is determinate. So the limit is 2.

Direct substitution yields \(\displaystyle{ \lim_{x \rightarrow 0}{\left[ \frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]} = \frac{-3}{0}}\). So the answer is \(+\infty\), \(-\infty\) or DNE (Does Not Exist). How do you know which one (without a graph)?
We need to determine the behavior of the function on either side of \(x=0\). On the left side of zero when \(x < 0\) and is very close to zero, the numerator is negative. We know this since x is very, very small, so \(x^4\) is positive but very small and \(5x\) is very small but negative. Both terms will be dominated by -3, so the numerator is negative. In the denominator, we need to look at the square root; \(x^2 \) is always positive and adding a positive number to 4 means that the number under the square root is always greater than 4, so the square root is always larger than 2. This makes the denominator negative. So we have a negative numerator and a negative denominator, so the limit \(\displaystyle{ \lim_{x \rightarrow 0^-}{\left[ \frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}= +\infty }\)
We use similar logic when \(x > 0\), to get \(\displaystyle{ \lim_{x \rightarrow 0^+}{\left[ \frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}= +\infty }\)

Since the limit from the right is equal to the limit from the left, the original limit is also \( +\infty \).

First of all, you always want to try direct substitution. When you do this, you get \(0/0\) which is indeterminate.
There are at least two ways to solve this. Since this is a ratio of polynomials, we will use algebra and factor. You can also use L'Hôpital's Rule.
First, we factor the numerator and denominator and cancel like terms, as follows:

\(\displaystyle{\lim_{x \rightarrow 2} \left[ \frac{3x^2-x-10}{x^2-4} \right] = \lim_{x \rightarrow 2} \left[ \frac{(x-2)(3x+5)}{(x-2)(x+2)} \right]}\)

Now the \((x-2)\) term that appears in both the numerator and denominator can cancel, giving

\(\displaystyle{\lim_{x \rightarrow 2} \left[ \frac{3x+5}{x+2} \right]}\)

Now let's try substitution again.

\(\displaystyle{\lim_{x \rightarrow 2} \left[ \frac{3x+5}{x+2} \right] = \frac{3(2)+5}{2+2} = \frac{11}{4} }\)

This technique will work only with polynomials that can be factored.

First of all, you always want to try direct substitution. When you do this in this limit, you get \(0/0\) which is indeterminate.
There are at least two ways to solve this. We will work it using algebra (since this is a ratio of polynomials). (You can also use L'Hôpital's Rule, since direct substitution yields an indeterminate form ).
First, we factor the numerator and denominator and cancel like terms.
\(\displaystyle{\lim_{x \rightarrow 3} \left[ \frac{x^4-81}{2x^2-5x-3} \right] = }\) \(\displaystyle{ \lim_{x \rightarrow 3} \left[ \frac{(x^2-9)(x^2+9)}{(x-3)(2x+1)} \right] = }\) \(\displaystyle{ \lim_{x \rightarrow 3} \left[ \frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)} \right]}\)
Now the \((x-3)\) term that appears in both the numerator and denominator can cancel, giving
\(\displaystyle{\lim_{x \rightarrow 3} \left[ \frac{(x+3)(x^2+9)}{(2x+1)} \right] }\)
Now let's try substitution again.
\(\displaystyle{ \lim_{x \rightarrow 3} \left[ \frac{(x+3)(x^2+9)}{(2x+1)} \right] = \frac{(3+3)(3^2+9)}{(2(3)+1)} = \frac{108}{7}}\).
This way will work with only polynomials and only if you can factor.

She made a mistake in notation here. In the last line of the first column, she wrote \(\displaystyle{ \lim_{x \to 1}{\frac{x+2}{x-3}}}\) which is correct. However, in the first line of the next column, she should have written \(\displaystyle{\frac{1+2}{1-3}=\frac{3}{-2}}\) WITHOUT the limit as x goes to 1 in front of it. Once you substitute the value, you do not write the limit.

The first thing you want to try here is to plug in -2 for t. However, in this limit, we get 0/0, which is an indeterminate form. So let's try factoring. But first we need to get the fraction in a different form.
In the numerator, we have \(\displaystyle{ \frac{1}{t} + \frac{1}{2} = \frac{2}{2t} + \frac{t}{2t} = \frac{t+2}{2t} }\).

So the limit looks like \(\displaystyle{\lim_{t \rightarrow -2}{ \left[ \frac{ \frac{t+2}{2t}}{t^3+8} \right]} = \lim_{t \rightarrow -2}{ \left[ \frac{t+2}{(2t)(t^3+8)} \right]} }\).

Now we need to factor the denominator. Let's check quickly to make sure (t+2) is a factor. This is a great time to practice synthetic division. Using synthetic division, \(t^3+8\) factors into \((t+2)(t^2-2t+4)\).
So our limit is now \(\displaystyle{\lim_{t \rightarrow -2}{ \left[ \frac{t+2}{(2t)(t+2)(t^2-2t+4)} \right]} }\). We can cancel the (t+2) term in the numerator and denominator and get \(\displaystyle{\lim_{t \rightarrow -2} \left[ \frac{1}{(2t)(t^2-2t+4)} \right]}\).
Substituting -2 for t gives us \(\displaystyle{ \frac{1}{(-4)((-2)^2-2(-2)+4)} = \frac{1}{-4(4+4+4)} = \frac{-1}{48} }\)

\(\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^3-7x}{x^3}\right]} = \lim_{x\rightarrow 0}{\left[ \frac{x(x^2-7)}{x(x^2)} \right]} = \lim_{x \rightarrow 0}{\left[ \frac{x^2-7}{x^2} \right]} }\)
The trick here is determine if the limit is \( +\infty \), \( -\infty \) or DNE (does not exist). So we need to look at the limit as x approaches zero from the left and compare it to the limit from the right. If they are different, then the limit does not exist.
Let's look at the left side of zero first. When x is less than zero (negative), \(x^2\) is always positive. So the numerator is negative and the denominator is positive telling us that \(\displaystyle{ \lim_{x \rightarrow 0^- }{\left[ \frac{x^2-7}{x^2} \right]}= -\infty }\).
Now, looking at the right side of zero, x is positive and \(x^2\) is also positive. Since the numerator is negative and the denominator is positive, we know that \(\displaystyle{ \lim_{x \rightarrow 0^+ }{\left[ \frac{x^2-7}{x^2} \right]}= -\infty }\).
Since \(\displaystyle{ \lim_{x \rightarrow 0^- }{\left[ \frac{x^2-7}{x^2} \right]}= \lim_{x \rightarrow 0^- }{\left[ \frac{x^2-7}{x^2} \right]}= -\infty }\), the limit \(\displaystyle{ \lim_{x \rightarrow 0 }{\left[ \frac{x^2-7}{x^2} \right]}= -\infty }\)

Direct substitution yields \(0/0\) which is indeterminate.
Factoring yields
\(\displaystyle{ \lim_{z \rightarrow 1}{\left[\frac{z^3-1}{(z-1)^2}\right]} = }\) \(\displaystyle{\lim_{z \rightarrow 1}{\left[\frac{(z-1)(z^2+z+1)}{(z-1)^2}\right]} = }\) \(\displaystyle{\lim_{z \rightarrow 1}{\left[\frac{z^2+z+1}{z-1}\right]}}\)
Direct substitution here gives us \(3/0\). So is the answer \(+\infty\) or \(-\infty\)? (without looking at a graph)
Well, looking at values close to 1, when \(z < 1\) the result is negative and when \(z>1\) the result is positive. So the answer is neither \(+\infty\) nor \(-\infty\) and the only correct is answer is that the limit does not exist.

Of course, the first thing we want to try is to plug in 27 directly for x. This gives us \(0/0\) which is indeterminate. So we need to do something else.
Upon initial inspection, this expression doesn't seem to be factorable. However, I am going to show you a trick. I don't know about you but I am not completely comfortable working with fractional powers. So I am going to substitute \(t\) for \(x^{1/3}\). We can do that as long we do the substitution for every expression containing x. So what do we get?

\(\displaystyle{ \frac{x-27}{x^{1/3}-3} = \frac{t^3-27}{t-3} }\)

Notice that since \(t=x^{1/3}\), we can cube both sides to get \(t^3=x\).

Using synthetic division, we can factor the numerator which becomes \((t^3-27)=(t-3)(t^2+3t+9)\).

Now the expression is \(\displaystyle{ \frac{(t-3)(t^2+3t+9)}{t-3}}\) and we can easily see that the \(t-3\) will cancel.
So let's go back to x's since our limit is in terms of x.

\(\displaystyle{ \begin{array}{rcl} \lim_{x \rightarrow 27}{\frac{x-27}{x^{1/3}-3}} & = & \lim_{x \rightarrow 27}{\frac{(x^{1/3}-3)(x^{2/3}+3x^{1/3}+9)}{x^{1/3}-3}} \\ & = & \lim_{x \rightarrow 27}{\frac{(x^{2/3}+3x^{1/3}+9)}{1}} \\ & = & 27^{2/3}+3(27)^{1/3}+9 = 27 \end{array} }\)

Note - - We could have used L'Hôpital's Rule here, which would have been much easier.

Direct substitution yields \(0/0\), which is indeterminate. To solve this exercise we need to do a trick.
We want to see if we can factor this problem to cancel terms in the numerator and denominator. I don't know about you but I have a hard time working with fractional powers. So I am going to use substitution to remove fractional powers.
In this problem, I am going to substitute for x. I want to get rid of both the 1/3rd and 1/4th powers. So, to do that I am going to let \(t=x^{1/12}\). I chose 1/12 since 1/12=(1/3)(1/4). You will see in a minute why this works.
Okay, so we have \(t=x^{1/12}\) which means \(t^{12}=x\) (by taking the 12th power of both sides).
Now, let's substitute.
\(\displaystyle{ \frac{x^{1/3}-1}{x^{1/4}-1} = \frac{(t^{12})^{1/3}-1}{(t^{12})^{1/4}-1} = \frac{t^4-1}{t^3-1} }\)
Okay, so I can factor the numerator and denominator here (use synthetic division) and I get
\(\displaystyle{ \frac{(t-1)(t^3+t^2+t+1)}{(t-1)(t^2+t+1)}}\)
Even though this is not completely factored, we can see that we can cancel (t-1) in this fraction, so we can stop factoring here.
Okay, let's go back to limit notation and convert back to x's.
\(\displaystyle{ \lim_{x \rightarrow 1}{\frac{(x^{1/12}-1)(x^{1/4}+x^{1/6}+x^{1/12}+1)}{(x^{1/12}-1)(x^{1/6}+x^{1/12}+1)}}=}\) \(\displaystyle{\lim_{x \rightarrow 1}{\frac{x^{1/4}+x^{1/6}+x^{1/12}+1}{x^{1/6}+x^{1/12}+1}}}\)
Now when we substitute 1 for x, we get \(\displaystyle{ \frac{1+1+1+1}{1+1+1} = 4/3 }\)

Note - - We could have used L'Hôpital's Rule here, which would have been much easier.

As usual, you want to plug in 4 for y and see what you get. In this limit, you get 0/0, which is indeterminate. Our next idea might be to factor. However, there doesn't appear to be anything we can factor here. So, we have to do something called rationalizing.
To do this, we take the term with the square root in it and multiply the numerator and denominator by its conjugate. In this case, the conjugate of \(3-\sqrt{y+5}\) is \(3+\sqrt{y+5}\). The conjugate is obtained by changing the sign between the terms.
So the limit becomes
\(\displaystyle{\lim_{y \rightarrow 4}{\frac{3-\sqrt{y+5}}{y-4}}=\lim_{y\rightarrow 4}{\left[\frac{3-\sqrt{y+5}}{y-4} \right] \left[ \frac{3+\sqrt{y+5}}{3+\sqrt{y+5}} \right] } }\)
Keep in mind that we need to multiply the numerator and denominator by the SAME EXPRESSION. In essence, this means we are multiplying by 1. Otherwise we change the problem (which we don't want to do).
Okay, so now what do we do? Now we multiply out the numerator and leave the denominator factored (you will see why in a minute).
The numerator is \(\displaystyle{ (3-\sqrt{y+5})(3+\sqrt{y+5}) = 9 - 3\sqrt{y+5} + 3\sqrt{y+5} - (y+5) }\). Do you see where all those terms came from?
Now, notice that there are two terms that look like \(3\sqrt{y+5}\) with different signs. So they cancel each other. So we are left with \(9-(y+5) = 9-y-5 = 4-y\). That's what the numerator reduces to. Our limit now looks like this.
\(\displaystyle{ \lim_{y \rightarrow 4}{ \left[ \frac{4-y}{(y-4)(3+\sqrt{y+5})} \right] } }\)
We are almost there. It looks like we have a term to cancel in the numerator and denominator but something is different. One of them is \(4-y\) and the other is \(y-4\). So they are slightly different. We need to do one more thing before we can cancel (the terms have to be EXACTLY the same to cancel). We need to get \(4-y\) to be \(y-4\). Well, what if we just switch the terms. If we do that we get \(-y+4\) and that is obviously not \(y-4\). (Remember, the sign in front of each term 'sticks' to that term, so when we switch the terms, we need to keep the negative sign with the y.)
To change \((-y+4)\) to \((y-4)\) we need to factor out a -1; (-y+4) = -1(y-4) giving us
\(\displaystyle{ \lim_{y \rightarrow 4}{ \left[ \frac{-1(y-4)}{(y-4)(3+\sqrt{y+5})} \right] } }\).
Now you can see why we didn't want to multiply out the denominator. If we had, it would not have been obvious that we can cancel the \((y-4)\) in numerator and denominator. Doing so we get
\(\displaystyle{ \lim_{y \rightarrow 4}{ \left[ \frac{-1}{(3+\sqrt{y+5})} \right] } }\) and substituting 4 for y gives us \(\displaystyle{ \frac{-1}{3+\sqrt{4+5}} = -1/6 }\).

Note:
This technique works whether the square root term is in the numerator or denominator and it doesn't matter if the square root term appears first or second. Just change one of the signs and not the other.

She makes some notation mistakes in this video. Make sure you do not make the same mistakes.