Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{\sin(3x)}{x}}}\)

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{\sin(5x)}{x}}}\)

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

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{t\to0}{\frac{\sin(5t)}{3t}}}\)

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:

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

Evaluate the limit \(\displaystyle{\lim_{\theta\to0}{\frac{\sin(5\theta)}{2\theta}}}\) giving your answer in exact terms.

This limit looks a lot like \(\displaystyle{ \lim_{\theta\to0}{ \frac{\sin(\theta)}{\theta} } }\). However, instead of just \(\theta\) in the sine term, we have \(5\theta\). This is okay as long as we have the same \(5\theta\) in the denominator, but we don't. We have \(2\theta\). So, first, let's factor out \(1/2\).

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

Evaluate the limit \(\displaystyle{\lim_{x\to0}{\frac{\sin^2(x)}{x^2}}}\) giving your answer in exact terms.

Direct substitution yields \(0/0\) which is indeterminate. So we need to find another way. The limit looks a lot like \(\displaystyle{ \lim_{\theta\to0}{ \frac{\sin(\theta)}{\theta} } }\). However, our limit has squared terms. But, we can factor to get

\(\displaystyle{\lim_{x\to0}{\frac{\sin^2(x)}{x^2}}}\) \( = 1 \)

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{\theta\to0}{\frac{\theta^2}{\sin(\theta)}}}\)

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{t\to0}{\frac{\sin^2(3t)}{t^2}}}\)

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{\sin(x)}{\sqrt{x}}}}\)

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

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{1}{x}\sin(x/3)}}\)

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

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{1-\cos(x)}{\sin(x)}}}\)

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.

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{[x \cdot \sec(x)\csc(x)]}}\)

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\pi/4}{\frac{\sin(x)+\cos(x)}{\tan(x)}}}\)

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\pi}{\sin(x+\sin(x))}}\)

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{\theta\to0}{\frac{\cos(\theta)-1}{\sin(\theta)}}}\)

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{\theta\to0}{\frac{\sin(\cos\theta)}{\sec(\theta)}}}\)

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)-1}{\cos x-1}}}\)

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

\(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)-1}{\cos x-1}}=4}\)

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{\theta\to1/2}{[\theta \cdot \sec(\pi\theta)]}}\)

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.

\(\displaystyle{\lim_{\theta\to1/2}{[\theta\sec(\pi\theta)]}}\) does not exist

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{t\to0}{\frac{\tan(6t)}{\sin(2t)}}}\)

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{\theta\to0}{\frac{1-\cos(\theta)}{\theta^2}}}\)

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

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\pi/4}{\frac{\sin(x)-\cos(x)}{\cos(2x)}}}\)

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to1}{\frac{\sin(x-1)}{x^2+x-2}}}\)

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{t\to0}{\frac{\tan(5t)}{\sin(2t)}}}\)

Evaluate the limit \(\displaystyle{\lim_{x\to0}{\frac{\cot(2x)}{\csc(x)}}}\) giving your answer in exact terms.

This problem has cotangent and cosecant. I can more easily work with sine and cosine. So, first, let's convert the function to sines and cosines using trig identities so see if we can simplify the function.

The last few simplifications were not necessary in order to complete the problem. In fact, once the sine terms were canceled, the resulting limit was determinate.
\(\displaystyle{\lim_{x\to0}{ \left[ \frac{\cos(2x)}{2 \cos x} \right] } = \frac{\cos(0)}{2\cos(0)} = \frac{1}{2} }\).
On your exam, you may want to try substitution to test for a determinate form before doing too much work. This will save you some time.

\(\displaystyle{\lim_{x\to0}{\frac{\cot(2x)}{\csc(x)}}}\) \( = 1/2 \)