\(\displaystyle{\lim_{t\to0}{\frac{\sin(5t)}{3t}}}\)

\(\displaystyle{\lim_{t\to0}{\frac{\sin(5t)}{3t}}}\)

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:

\(\displaystyle{\lim_{x\to0}{\frac{\sin(3x)}{x}}}\)

\(\displaystyle{\lim_{x\to0}{\frac{\cot(2x)}{\csc(x)}}}\)

\(\displaystyle{\lim_{x\to0}{\frac{\sin(5x)}{2x}}}\)

\(\displaystyle{\lim_{x\to0}{\frac{\sin(5x)}{2x}}}\)

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

\(\displaystyle{\lim_{x\to0}{\frac{\sin^2(x)}{x^2}}}\)

\(\displaystyle{\lim_{\theta\to0}{\frac{\theta^2}{\sin(\theta)}}}\)

\(\displaystyle{\lim_{x\to0}{\frac{\sin(5x)}{x}}}\)

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

\(\displaystyle{\lim_{x\to0}{\frac{\sin(x)}{\sqrt{x}}}}\)

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

\(\displaystyle{\lim_{x\to0}{\frac{1}{x}\sin(x/3)}}\)

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

\(\displaystyle{\lim_{x\to0}{\frac{1-\cos(x)}{\sin(x)}}}\)

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.

\(\displaystyle{\lim_{x\to0}{x\sec(x)\csc(x)}}\)

\(\displaystyle{\lim_{x\to\pi/4}{\frac{\sin(x)+\cos(x)}{\tan(x)}}}\)

\(\displaystyle{\lim_{x\to\pi}{\sin(x+\sin(x))}}\)

\(\displaystyle{\lim_{\theta\to0}{\frac{\cos(\theta)-1}{\sin(\theta)}}}\)

\(\displaystyle{\lim_{\theta\to0}{\frac{\sin(\cos\theta)}{\sec(\theta)}}}\)

\(\displaystyle{\lim_{t\to0}{\frac{\sin^2(3t)}{t^2}}}\)

\(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)-1}{\cos x-1}}}\)

\(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)-1}{\cos x-1}}}\)

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

\(\displaystyle{\lim_{\theta\to1/2}{[\theta\sec(\pi\theta)]}}\)

\(\displaystyle{\lim_{\theta\to1/2}{[\theta\sec(\pi\theta)]}}\)

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 debatable and is different from how we define DNE (does not exist) 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.

\(\displaystyle{\lim_{t\to0}{\frac{\tan(6t)}{\sin(2t)}}}\)

\(\displaystyle{\lim_{\theta\to0}{\frac{1-\cos(\theta)}{\theta^2}}}\)

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

\(\displaystyle{\lim_{x\to\pi/4}{\frac{\sin(x)-\cos(x)}{\cos(2x)}}}\)

\(\displaystyle{\lim_{x\to1}{\frac{\sin(x-1)}{x^2+x-2}}}\)

\(\displaystyle{\lim_{t\to0}{\frac{\tan(5t)}{\sin(2t)}}}\)