17Calculus - Limits Involving Trig Functions
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.
Dr Chris Tisdell - Beginners Guide to Special Trig Limits in Calculus
video by Dr Chris Tisdell |
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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.
Practice
Unless otherwise instructed, evaluate these limits. Give your answers in exact terms.
Basic |
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\(\displaystyle{\lim_{t\to0}{\frac{\sin(5t)}{3t}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{t\to0}{\frac{\sin(5t)}{3t}}}\)
Final Answer |
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\(\displaystyle{\lim_{t\to0}{\frac{\sin(5t)}{3t}}=5/3}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{t\to0}{\frac{\sin(5t)}{3t}}}\)
Solution |
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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 |
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\(\displaystyle{\lim_{t\to0}{\frac{\sin(5t)}{3t}}=5/3}\) |
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\(\displaystyle{\lim_{x\to0}{\frac{\sin(3x)}{x}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{\sin(3x)}{x}}}\)
Solution |
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28 video
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\(\displaystyle{\lim_{\theta\to0}{\frac{\theta^2}{\sin(\theta)}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{\theta\to0}{\frac{\theta^2}{\sin(\theta)}}}\)
Solution |
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451 video
video by Krista King Math |
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\(\displaystyle{\lim_{x\to0}{\frac{\sin(5x)}{x}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{\sin(5x)}{x}}}\)
Solution |
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Notice that she drops the limit notation in her work, which could cost you points on your homework or exam.
453 video
video by Krista King Math |
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\(\displaystyle{\lim_{x\to0}{\frac{\sin(x)}{\sqrt{x}}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{\sin(x)}{\sqrt{x}}}}\)
Solution |
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Notice that she drops the limit notation in her work, which could cost you points on your homework or exam.
454 video
video by Krista King Math |
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\(\displaystyle{\lim_{x\to0}{\frac{1}{x}\sin(x/3)}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{1}{x}\sin(x/3)}}\)
Solution |
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Notice that she drops the limit notation in her work, which could cost you points on your homework or exam.
455 video
video by Krista King Math |
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\(\displaystyle{\lim_{x\to0}{\frac{1-\cos(x)}{\sin(x)}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{1-\cos(x)}{\sin(x)}}}\)
Solution |
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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.
456 video
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\(\displaystyle{\lim_{x\to0}{[x \cdot \sec(x)\csc(x)]}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{[x \cdot \sec(x)\csc(x)]}}\)
Solution |
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457 video
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\(\displaystyle{\lim_{x\to\pi/4}{\frac{\sin(x)+\cos(x)}{\tan(x)}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\pi/4}{\frac{\sin(x)+\cos(x)}{\tan(x)}}}\)
Solution |
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458 video
video by PatrickJMT |
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\(\displaystyle{\lim_{x\to\pi}{\sin(x+\sin(x))}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\pi}{\sin(x+\sin(x))}}\)
Solution |
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459 video
video by PatrickJMT |
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\(\displaystyle{\lim_{\theta\to0}{\frac{\cos(\theta)-1}{\sin(\theta)}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{\theta\to0}{\frac{\cos(\theta)-1}{\sin(\theta)}}}\)
Solution |
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460 video
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\(\displaystyle{\lim_{\theta\to0}{\frac{\sin(\cos\theta)}{\sec(\theta)}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{\theta\to0}{\frac{\sin(\cos\theta)}{\sec(\theta)}}}\)
Solution |
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461 video
video by PatrickJMT |
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\(\displaystyle{\lim_{t\to0}{\frac{\sin^2(3t)}{t^2}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{t\to0}{\frac{\sin^2(3t)}{t^2}}}\)
Solution |
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463 video
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\(\displaystyle{\lim_{x\to0}{\frac{\cot(2x)}{\csc(x)}}}\)
Problem Statement |
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valuate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{\cot(2x)}{\csc(x)}}}\)
Solution |
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31 video
video by Khan Academy |
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\(\displaystyle{\lim_{x\to0}{\frac{\sin(5x)}{2x}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms.\(\displaystyle{\lim_{x\to0}{\frac{\sin(5x)}{2x}}}\)
Final Answer |
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\(\displaystyle{\lim_{x\to0}{\frac{\sin(5x)}{2x}}=\frac{5}{2}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms.\(\displaystyle{\lim_{x\to0}{\frac{\sin(5x)}{2x}}}\)
Solution |
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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 \(\displaystyle{\lim_{x\to0}{\frac{5}{2}\frac{\sin(5x)}{5x}}=\frac{5}{2}(1)=\frac{5}{2}}\)
35 video
video by Khan Academy |
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Final Answer |
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\(\displaystyle{\lim_{x\to0}{\frac{\sin(5x)}{2x}}=\frac{5}{2}}\) |
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\(\displaystyle{\lim_{x\to0}{\frac{\sin^2(x)}{x^2}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{\sin^2(x)}{x^2}}}\)
Solution |
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36 video
video by Khan Academy |
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Intermediate |
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\(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)-1}{\cos x-1}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)-1}{\cos x-1}}}\)
Final Answer |
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\(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)-1}{\cos x-1}}=4}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)-1}{\cos x-1}}}\)
Solution |
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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 |
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\(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)-1}{\cos x-1}}=4}\) |
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\(\displaystyle{\lim_{\theta\to1/2}{[\theta \cdot \sec(\pi\theta)]}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{\theta\to1/2}{[\theta \cdot \sec(\pi\theta)]}}\)
Final Answer |
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\(\displaystyle{\lim_{\theta\to1/2}{[\theta\sec(\pi\theta)]}}\) does not exist
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{\theta\to1/2}{[\theta \cdot \sec(\pi\theta)]}}\)
Solution |
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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.
21 video
video by Krista King Math |
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Final Answer |
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\(\displaystyle{\lim_{\theta\to1/2}{[\theta\sec(\pi\theta)]}}\) does not exist |
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\(\displaystyle{\lim_{t\to0}{\frac{\tan(6t)}{\sin(2t)}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{t\to0}{\frac{\tan(6t)}{\sin(2t)}}}\)
Solution |
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29 video
video by PatrickJMT |
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\(\displaystyle{\lim_{\theta\to0}{\frac{1-\cos(\theta)}{\theta^2}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{\theta\to0}{\frac{1-\cos(\theta)}{\theta^2}}}\)
Solution |
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To save space, she dropped the limit notation in her work. Make sure you don't do that on your homework or exam.
452 video
video by Krista King Math |
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\(\displaystyle{\lim_{x\to\pi/4}{\frac{\sin(x)-\cos(x)}{\cos(2x)}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to\pi/4}{\frac{\sin(x)-\cos(x)}{\cos(2x)}}}\)
Solution |
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462 video
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\(\displaystyle{\lim_{x\to1}{\frac{\sin(x-1)}{x^2+x-2}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{x\to1}{\frac{\sin(x-1)}{x^2+x-2}}}\)
Solution |
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464 video
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\(\displaystyle{\lim_{t\to0}{\frac{\tan(5t)}{\sin(2t)}}}\)
Problem Statement |
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Evaluate this limit, giving your answer in exact terms. \(\displaystyle{\lim_{t\to0}{\frac{\tan(5t)}{\sin(2t)}}}\)
Solution |
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465 video
video by PatrickJMT |
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trig limits 17calculus youtube playlist
Here is a playlist of the videos on this page.
You CAN Ace Calculus
Topics You Need To Understand For This Page
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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Practice Instructions
Unless otherwise instructed, evaluate these limits. Give your answers in exact terms.
