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.
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 \(3\theta\).
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.
Instructions  Unless otherwise instructed, evaluate these limits. Give your answers in exact terms.
Basic Problems 

Problem Statement 

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

Problem Statement 

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

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

Log in to rate this practice problem. 

Problem Statement 

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

video by PatrickJMT 

close solution

Log in to rate this practice problem. 

Problem Statement 

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

video by Khan Academy 

close solution

Log in to rate this practice problem. 

Problem Statement 

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

Problem Statement 

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

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 \(\displaystyle{\lim_{x\to0}{\frac{5}{2}\frac{\sin(5x)}{5x}}=\frac{5}{2}(1)=\frac{5}{2}}\)
video by Khan Academy 

Final Answer 

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

Log in to rate this practice problem. 

Problem Statement 

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

video by Khan Academy 

close solution

Log in to rate this practice problem. 

Problem Statement 

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

video by Krista King Math 

close solution

Log in to rate this practice problem. 

Problem Statement 

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

close solution

Log in to rate this practice problem. 

Problem Statement 

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

close solution

Log in to rate this practice problem. 

Problem Statement 

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

close solution

Log in to rate this practice problem. 

Problem Statement 

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

close solution

Log in to rate this practice problem. 

Problem Statement 

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

video by Krista King Math 

close solution

Log in to rate this practice problem. 

Problem Statement 

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

video by PatrickJMT 

close solution

Log in to rate this practice problem. 

Problem Statement 

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

video by PatrickJMT 

close solution

Log in to rate this practice problem. 

Problem Statement 

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

video by PatrickJMT 

close solution

Log in to rate this practice problem. 

Problem Statement 

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

video by PatrickJMT 

close solution

Log in to rate this practice problem. 

Problem Statement 

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

video by PatrickJMT 

close solution

Log in to rate this practice problem. 

Intermediate Problems 

Problem Statement 

\(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)1}{\cos x1}}}\)
Final Answer 

Problem Statement 

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

\(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)1}{\cos x1}}=4}\) 
close solution

Log in to rate this practice problem. 

Problem Statement 

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

Problem Statement 

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

Log in to rate this practice problem. 

Problem Statement 

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

video by PatrickJMT 

close solution

Log in to rate this practice problem. 

Problem Statement 

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

close solution

Log in to rate this practice problem. 

Problem Statement 

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

video by PatrickJMT 

close solution

Log in to rate this practice problem. 

Problem Statement 

\(\displaystyle{\lim_{x\to1}{\frac{\sin(x1)}{x^2+x2}}}\)
Solution 

video by PatrickJMT 

close solution

Log in to rate this practice problem. 

Problem Statement 

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

video by PatrickJMT 

close solution

Log in to rate this practice problem. 

Here is a playlist of the videos on this page.
You CAN Ace Calculus
To bookmark this page and practice problems, log in to your account or set up a free account.
Single Variable Calculus 

MultiVariable Calculus 

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
 
Help Keep 17Calculus Free 

The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free. 