For this problem, you need to know this limit:
\(\displaystyle{ \lim_{h \to 0}{\frac{\sin( h )}{h}} = 1 }\)
Now, the key is that the 'h' terms need to be exactly the same, as shown in this limit.
However, our problem has a \(5t\) in the sine term and \(3t\) in the denominator. There's nothing we can do with the \(5t\) in the numerator term \(\sin(5t)\) but we can get \(5t\) in the denominator, like this:

\(\displaystyle{ \left(\frac{1}{3}\right)\frac{\sin(5t)}{t} = }\) \(\displaystyle{ \left(\frac{1}{3}\right)\frac{\sin(5t)}{t}\left(\frac{5}{5}\right) = }\) \(\displaystyle{ \left(\frac{5}{3}\right)\frac{\sin(5t)}{5t} }\)

Now, we can use the limit above where \(h=5t\) and we get

\(\displaystyle{ \lim_{t \to 0}{\frac{\sin(5t)}{3t}} = }\) \(\displaystyle{ \lim_{t \to 0}{\frac{5}{3}\frac{\sin(5t)}{5t}} = }\) \(\displaystyle{ \frac{5}{3}\lim_{t \to 0}{\frac{\sin(5t)}{5t}} = 5/3 }\)

He works this the hard way and it is not really recommended to do it this way. It is easier to factor out the \(1/2\) and then multiply the numerator and denominator by 5. This gives you \(\displaystyle{\lim_{x\to0}{\frac{5}{2}\frac{\sin(5x)}{5x}}=\frac{5}{2}(1)=\frac{5}{2}}\)

Notice that she drops the limit notation in her work, which could cost you points on your homework or exam.

Notice that she drops the limit notation in her work, which could cost you points on your homework or exam.

Notice that she drops the limit notation in her work, which could cost you points on your homework or exam.

She does this the hard way using conjugate multiplication. It is easier to just multiply the numerator and denominator by x and use the basic trig limit identities.

Direct evaluation yields \(0/0\), which is indeterminate. So we need to do something else. Notice we can't do anything with the \(\cos(2x)\) and \(\cos(x)\) terms. They are completely different.
This is a good example of where you want to try to use a trig identity.
Looking at the denominator, we want something in the numerator with a \(\cos(x)\) in it. So, try the identity \(\cos(2x) = 2 \cos^2 x - 1\).
Substituting this, we get
\(\displaystyle{ \begin{array}{rcl} & & \lim_{x \to 0}{\frac{\cos(2x) - 1}{\cos x - 1}} \\\\ & = & \lim_{x \to 0}{\frac{(2 \cos^2 x - 1) - 1}{\cos x - 1}} \\\\ & = & \lim_{x \to 0}{\frac{2 \cos^2 x - 2}{\cos x - 1}} \\\\ & = & \lim_{x \to 0}{\frac{2 ( \cos^2 x - 1) }{\cos x - 1}} \end{array} }\)

Factoring the numerator, we get

\(\displaystyle{ \lim_{x \to 0}{\frac{2 (\cos x - 1)(\cos x + 1) }{\cos x - 1}} = \lim_{x \to 0}{\frac{2(\cos x + 1) }{1}} }\)

Direct substitution now gives us \(2(1 + 1) = 4\)

1. In this video, she equates the limit going to infinity ( or negative infinity ) as a limit that does not exist. You know from the discussion on the main limits page that this is disputable and is different from how we use the idea on this site. Check with your instructor to see what they mean by the limit not existing. That said, we do agree with the final answer to this problem, since the limit from the left is \(+\infty\) and the limit from the right is \(-\infty\) and, since the limit from the right does not equal the limit from the left, the limit does not exist.
2. The title of this video and the comments below use the term 'infinite' to describe this limit. We use the term 'finite' to describe this limit since the x-value is approaching a finite value.

IntegralCALC Comments: Learn how to calculate an infinite limit. In order to determine whether or not the function has an infinite limit as x approaches a specific point, you'll need to first confirm that your function as a vertical asymptote and the point you're approaching. The function will have a vertical asymptote where it is undefined because the denominator of the function is equal to zero. Then you'll need to plug in values close to and on both sides of the number you're approaching. If the value of the function gets very large as you approach the vertical asymptote, then the limit is positive infinity. If the values get very large but negative as you approach the vertical asymptote, then the limit is negative infinity. Remember that infinite limits don't technically exist, but because it gives us so much more information about the function to say that the limit is positive of negative infinity, we'll often provide that answer instead of stating that the limit does not exist (DNE). If the one-sided limits are not equal, there is no general limit.

To save space, she dropped the limit notation in her work. Make sure you understand that you can't do that on your homework or exam.