When we are asked to determine a limit involving trig functions, the best strategy is always to try L'Hôpital's Rule. However, this rule is usually not covered until second semester calculus. So, to evaluate trig limits without L'Hôpital's Rule, we use the following identities.
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Trig Limit Identities
\(\displaystyle{ \lim_{\theta \to 0}{\frac{\sin(\theta)}{\theta}} = \lim_{\theta \to 0}{\frac{\theta}{\sin(\theta)}} = 1 }\) 
\(\displaystyle{ \lim_{\theta \to 0}{\frac{1\cos(\theta)}{\theta}} = \lim_{\theta \to 0}{\frac{\cos(\theta)1}{\theta}} = 0 }\) 
Before we get into the details of these identities, let's watch a quick video.
video by Dr Chris Tisdell 

Look over both of those limits carefully. Notice what \(\theta\) goes to. Also notice that the expression in the denominator must match the expression within the trig functions. So, for example, if you have \( \sin(3\theta)\) in the first limit, the denominator must also be \(3\theta\). Here is another video that talks more about the sine limit.
video by Trefor Bazett 

Steps To Evaluate Trig Limits
Step 1 [ direct substitution ]   directly substitute the variable into the trig function; if you get an indeterminate form, more work is required; if you don't, you are done.
Step 2A [ algebra ]   if you have an indeterminate form from direct substitution, use algebra to try to get your limit into a form that matches one or both identities above.
Step 2B [ trig identities ]   if you can't get your limit to match one of the identities above, use trig identities to get your limit into another form; you may be able to get cancellation or you may be able to match one or both of the identities above.
Step 3 [ keep trying ]   use direct substitution again to see if you no longer have an indeterminate form; you may need to use the Multiplication Rule when evaluating; if you still have an indeterminate form, don't give up; keep working with it until you get it.
Unless otherwise instructed, evaluate these limits. Give your answers in exact terms.
Basic
\(\displaystyle{\lim_{x\to0}{\frac{\sin(3x)}{x}}}\)
Problem Statement
Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{\sin(3x)}{x}}}\)
Solution
video by PatrickJMT 

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\(\displaystyle{\lim_{x\to0}{\frac{\sin(5x)}{x}}}\)
Problem Statement
Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{\sin(5x)}{x}}}\)
Solution
Notice that she drops the limit notation in her work, which could cost you points on your homework or exam.
video by Krista King Math 

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

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

\(\displaystyle{\lim_{t\to0}{\frac{\sin(5t)}{3t}}=5/3}\)
Problem Statement
Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{t\to0}{\frac{\sin(5t)}{3t}}}\)
Solution
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 \(h=5t\). 
\(\displaystyle{ \lim_{t \to 0}{\frac{\sin(5t)}{3t}} }\) 
\(\displaystyle{\lim_{t \to 0}{\frac{5}{3}\frac{\sin(5t)}{5t}} }\) 
\(\displaystyle{\frac{5}{3}\lim_{t \to 0}{\frac{\sin(5t)}{5t}} = \frac{5}{3}(1) }\) 
Final Answer
\(\displaystyle{\lim_{t\to0}{\frac{\sin(5t)}{3t}}=5/3}\)
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\(\displaystyle{\lim_{\theta\to0}{\frac{\sin(5\theta)}{2\theta}}}\)
Problem Statement 

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

\(\displaystyle{\lim_{\theta\to0}{\frac{\sin(5\theta)}{2\theta}}}\) \(\displaystyle{ = \frac{5}{2} }\)
Problem Statement
Evaluate the limit \(\displaystyle{\lim_{\theta\to0}{\frac{\sin(5\theta)}{2\theta}}}\) giving your answer in exact terms.
Solution
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 \(1/2\).
Using the Constant Multiple Rule, we can factor a \(2\) out of the denominator. 
\(\displaystyle{ \frac{1}{2} \lim_{\theta\to0}{\frac{\sin(5\theta)}{\theta}}}\) 
In order to use the sine limit, we need the sine term to be the same as the denominator. We can't do anything about the \(5\theta\) in the sine term but we can multiply the denominator by \(5\) to get \(5\theta\) in the denominator. However, in order to balance this equation and to not change the problem, we also need to multiply the numerator by \(5\). 
\(\displaystyle{ \frac{1}{2} \lim_{\theta\to0}{\frac{5\sin(5\theta)}{5\theta}}}\) 
Using the Constant Multiple Rule again, we have 
\(\displaystyle{ \frac{5}{2} \lim_{\theta\to0}{\frac{\sin(5\theta)}{5\theta}}}\) 
Now we can apply the sine limit to get the final answer. 
\(\displaystyle{ \frac{5}{2}(1) = \frac{5}{2} }\) 
Final Answer
\(\displaystyle{\lim_{\theta\to0}{\frac{\sin(5\theta)}{2\theta}}}\) \(\displaystyle{ = \frac{5}{2} }\)
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\(\displaystyle{\lim_{x\to0}{\frac{\sin^2(x)}{x^2}}}\)
Problem Statement 

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

\(\displaystyle{\lim_{x\to0}{\frac{\sin^2(x)}{x^2}}}\) \( = 1 \)
Problem Statement
Evaluate the limit \(\displaystyle{\lim_{x\to0}{\frac{\sin^2(x)}{x^2}}}\) giving your answer in exact terms.
Solution
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
\(\displaystyle{\lim_{x\to0}{ \left[ \frac{\sin(x)}{x} \frac{\sin(x)}{x} \right] } }\) 
We can apply the Multiplication Rule to get 
\(\displaystyle{ \lim_{x\to0}{ \left[ \frac{\sin(x)}{x} \right] } \cdot \lim_{x\to0}{ \left[ \frac{\sin(x)}{x} \right] } }\) 
Now we have two limits that match sine rule. Simplifying, we get 
\( 1 \cdot 1 = 1\) 
Another rule we could have applied is the Power Rule, as follows. 
\(\displaystyle{\lim_{x\to0}{\frac{\sin^2(x)}{x^2}}}\) \(\displaystyle{ = \lim_{x\to0}{ \left[ \frac{\sin(x)}{x} \right]^2 }}\) \(\displaystyle{ = \left[ \lim_{x\to0}{ \frac{\sin(x)}{x} } \right]^2 }\) \( = 1^2 = 1\) 
Final Answer
\(\displaystyle{\lim_{x\to0}{\frac{\sin^2(x)}{x^2}}}\) \( = 1 \)
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\(\displaystyle{\lim_{\theta\to0}{\frac{\theta^2}{\sin(\theta)}}}\)
Problem Statement
Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{\theta\to0}{\frac{\theta^2}{\sin(\theta)}}}\)
Solution
video by Krista King Math 

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

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\(\displaystyle{\lim_{x\to0}{\frac{\sin(x)}{\sqrt{x}}}}\)
Problem Statement
Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{\sin(x)}{\sqrt{x}}}}\)
Solution
Notice that she drops the limit notation in her work, which could cost you points on your homework or exam.
video by Krista King Math 

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\(\displaystyle{\lim_{x\to0}{\frac{1}{x}\sin(x/3)}}\)
Problem Statement
Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{1}{x}\sin(x/3)}}\)
Solution
Notice that she drops the limit notation in her work, which could cost you points on your homework or exam.
video by Krista King Math 

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\(\displaystyle{\lim_{x\to0}{\frac{1\cos(x)}{\sin(x)}}}\)
Problem Statement
Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{1\cos(x)}{\sin(x)}}}\)
Solution
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.
video by Krista King Math 

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\(\displaystyle{\lim_{x\to0}{[x \cdot \sec(x)\csc(x)]}}\)
Problem Statement
Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{[x \cdot \sec(x)\csc(x)]}}\)
Solution
video by Krista King Math 

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

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\(\displaystyle{\lim_{x\to\pi}{\sin(x+\sin(x))}}\)
Problem Statement
Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\pi}{\sin(x+\sin(x))}}\)
Solution
video by PatrickJMT 

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\(\displaystyle{\lim_{\theta\to0}{\frac{\cos(\theta)1}{\sin(\theta)}}}\)
Problem Statement
Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{\theta\to0}{\frac{\cos(\theta)1}{\sin(\theta)}}}\)
Solution
video by PatrickJMT 

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

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Intermediate
\(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)1}{\cos x1}}}\)
Problem Statement 

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)1}{\cos x1}}}\)
Final Answer 

\(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)1}{\cos x1}}=4}\)
Problem Statement
Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)1}{\cos x1}}}\)
Solution
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
\(\displaystyle{ \lim_{x \to 0}{\frac{\cos(2x)  1}{\cos x  1}} }\) 
\(\displaystyle{ \lim_{x \to 0}{\frac{(2 \cos^2 x  1)  1}{\cos x  1}} }\) 
\(\displaystyle{ \lim_{x \to 0}{\frac{2 \cos^2 x  2}{\cos x  1}} }\) 
\(\displaystyle{ \lim_{x \to 0}{\frac{2 ( \cos^2 x  1) }{\cos x  1}} }\) 
Factoring the numerator, we get 
\(\displaystyle{ \lim_{x \to 0}{\frac{2 (\cos x  1)(\cos x + 1) }{\cos x  1}} }\) 
\(\displaystyle{ \lim_{x \to 0}{\frac{2(\cos x + 1) }{1}} }\) 
Direct substitution now gives us 
\(\displaystyle{ \frac{2(1+1)}{1} = 4 }\) 
Final Answer
\(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)1}{\cos x1}}=4}\)
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\(\displaystyle{\lim_{\theta\to1/2}{[\theta \cdot \sec(\pi\theta)]}}\)
Problem Statement 

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

\(\displaystyle{\lim_{\theta\to1/2}{[\theta\sec(\pi\theta)]}}\) does not exist
Problem Statement
Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{\theta\to1/2}{[\theta \cdot \sec(\pi\theta)]}}\)
Solution
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.
video by Krista King Math 

Final Answer
\(\displaystyle{\lim_{\theta\to1/2}{[\theta\sec(\pi\theta)]}}\) does not exist
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\(\displaystyle{\lim_{t\to0}{\frac{\tan(6t)}{\sin(2t)}}}\)
Problem Statement
Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{t\to0}{\frac{\tan(6t)}{\sin(2t)}}}\)
Solution
video by PatrickJMT 

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\(\displaystyle{\lim_{\theta\to0}{\frac{1\cos(\theta)}{\theta^2}}}\)
Problem Statement
Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{\theta\to0}{\frac{1\cos(\theta)}{\theta^2}}}\)
Solution
To save space, she dropped the limit notation in her work. Make sure you don't do that on your homework or exam.
video by Krista King Math 

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\(\displaystyle{\lim_{x\to\pi/4}{\frac{\sin(x)\cos(x)}{\cos(2x)}}}\)
Problem Statement
Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\pi/4}{\frac{\sin(x)\cos(x)}{\cos(2x)}}}\)
Solution
video by PatrickJMT 

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\(\displaystyle{\lim_{x\to1}{\frac{\sin(x1)}{x^2+x2}}}\)
Problem Statement
Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to1}{\frac{\sin(x1)}{x^2+x2}}}\)
Solution
video by PatrickJMT 

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

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

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

\(\displaystyle{\lim_{x\to0}{\frac{\cot(2x)}{\csc(x)}}}\) \( = 1/2 \)
Problem Statement
Evaluate the limit \(\displaystyle{\lim_{x\to0}{\frac{\cot(2x)}{\csc(x)}}}\) giving your answer in exact terms.
Solution
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.
Here are some trig identities that we will use. 
\(\displaystyle{ \cot \theta = \frac{\cos \theta}{\sin \theta} }\) 
\(\displaystyle{ \csc \theta = \frac{1}{\sin \theta} }\) 
\( \sin(2\theta) = 2\sin \theta \cos \theta \) 
We will probably need an identity for \(\cos(2\theta)\) but there are several to choose from and, at this point, we are not sure which one to use. So let's use these and decide after that. 
\(\displaystyle{\lim_{x\to0}{\frac{\cot(2x)}{\csc(x)}}}\) 
Using the first two identities we have 
\(\displaystyle{\lim_{x\to0}{ \left[ \frac{\cos(2x)}{\sin(2x)} \cdot \frac{1}{1/\sin x} \right] } }\) 
\(\displaystyle{\lim_{x\to0}{ \left[ \frac{(\sin x)(\cos(2x))}{2\sin x \cos x} \right] } }\) 
\(\displaystyle{\lim_{x\to0}{ \left[ \frac{\cos(2x)}{2 \cos x} \right] } }\) 
At this point, it looks like the trig identity \( \cos(2\theta) = 2\cos^2 \theta  1 \) would be best to use here since then we can do some cancelation with the cosines. 
\(\displaystyle{\lim_{x\to0}{ \left[ \frac{2\cos^2 x  1}{2 \cos x} \right] } }\) 
Let's simplify one more time before we try to plug in the limit. 
\(\displaystyle{\lim_{x\to0}{ \left[ \frac{2\cos^2 x}{2 \cos x}  \frac{1}{2 \cos x} \right] } }\) 
\(\displaystyle{\lim_{x\to0}{ \left[ \cos x  \frac{1}{2 \cos x} \right] } }\) 
Now try plugging in \(x=0\) to see if the limit is determinate. 
\( \cos 0  1/(2\cos 0) = 1  1/2 = 1/2 \) 
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.
Final Answer
\(\displaystyle{\lim_{x\to0}{\frac{\cot(2x)}{\csc(x)}}}\) \( = 1/2 \)
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Really UNDERSTAND Calculus
Topics You Need To Understand For This Page 

[precalculus]  [basics of limits]  [finite limits]  [infinite limits] 
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Practice Instructions
Unless otherwise instructed, evaluate these limits. Give your answers in exact terms.