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Limits Involving Trig and Inverse Trig Functions 
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► trig limit identities ► steps to evaluate trig limits 
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 
\(\displaystyle{ \lim_{\theta \to 0}{\frac{\sin(\theta)}{\theta}} =
\lim_{\theta \to 0}{\frac{\theta}{\sin(\theta)}} = 1 }\) 

\(\displaystyle{ \lim_{\theta \to 0}{\frac{1\cos(\theta)}{\theta}} =
\lim_{\theta \to 0}{\frac{\cos(\theta)1}{\theta}} = 0 }\) 
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.
Practice A01 
\(\displaystyle{\lim_{t\to0}{\frac{\sin(5t)}{3t}}}\) 


\(\displaystyle{\lim_{t\to0}{\frac{\sin(5t)}{3t}}=5/3}\) 
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 \(h=5t\) and we get
\(\displaystyle{ \lim_{t \to 0}{\frac{\sin(5t)}{3t}} = }\) \(\displaystyle{
\lim_{t \to 0}{\frac{5}{3}\frac{\sin(5t)}{5t}} = }\) \(\displaystyle{
\frac{5}{3}\lim_{t \to 0}{\frac{\sin(5t)}{5t}} = 5/3 }\)
Practice A01 Final Answer 
\(\displaystyle{\lim_{t\to0}{\frac{\sin(5t)}{3t}}=5/3}\) 
Practice A02 
\(\displaystyle{\lim_{x\to0}{\frac{\sin(3x)}{x}}}\) 

Practice A03 
\(\displaystyle{\lim_{x\to0}{\frac{\cot(2x)}{\csc(x)}}}\) 

Practice A04 
\(\displaystyle{\lim_{x\to0}{\frac{\sin(5x)}{2x}}}\) 


\(\displaystyle{\lim_{x\to0}{\frac{\sin(5x)}{2x}}=\frac{5}{2}}\) 
He works this the hard way and it is not really recommended to do it this way. It is easier to factor out the \(1/2\) and then multiply the numerator and denominator by 5. This gives you \(\displaystyle{\lim_{x\to0}{\frac{5}{2}\frac{\sin(5x)}{5x}}=\frac{5}{2}(1)=\frac{5}{2}}\)
Practice A04 Final Answer 
\(\displaystyle{\lim_{x\to0}{\frac{\sin(5x)}{2x}}=\frac{5}{2}}\) 
Practice A05 
\(\displaystyle{\lim_{x\to0}{\frac{\sin^2(x)}{x^2}}}\) 

Practice A06 
\(\displaystyle{\lim_{\theta\to0}{\frac{\theta^2}{\sin(\theta)}}}\) 

Practice A07 
\(\displaystyle{\lim_{x\to0}{\frac{\sin(5x)}{x}}}\) 

Notice that she drops the limit notation in her work, which could cost you points on your homework or exam.
Practice A08 
\(\displaystyle{\lim_{x\to0}{\frac{\sin(x)}{\sqrt{x}}}}\) 

Notice that she drops the limit notation in her work, which could cost you points on your homework or exam.
Practice A09 
\(\displaystyle{\lim_{x\to0}{\frac{1}{x}\sin(x/3)}}\) 

Notice that she drops the limit notation in her work, which could cost you points on your homework or exam.
Practice A10 
\(\displaystyle{\lim_{x\to0}{\frac{1\cos(x)}{\sin(x)}}}\) 

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.
Practice A11 
\(\displaystyle{\lim_{x\to0}{x\sec(x)\csc(x)}}\) 

Practice A12 
\(\displaystyle{\lim_{x\to\pi/4}{\frac{\sin(x)+\cos(x)}{\tan(x)}}}\) 

Practice A13 
\(\displaystyle{\lim_{x\to\pi}{\sin(x+\sin(x))}}\) 

Practice A14 
\(\displaystyle{\lim_{\theta\to0}{\frac{\cos(\theta)1}{\sin(\theta)}}}\) 

Practice A15 
\(\displaystyle{\lim_{\theta\to0}{\frac{\sin(\cos\theta)}{\sec(\theta)}}}\) 

Practice A16 
\(\displaystyle{\lim_{t\to0}{\frac{\sin^2(3t)}{t^2}}}\) 

Practice B01 
\(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)1}{\cos x1}}}\) 


\(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)1}{\cos x1}}=4}\) 
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
\(\displaystyle{
\begin{array}{rcl}
& & \lim_{x \to 0}{\frac{\cos(2x)  1}{\cos x  1}} \\\\
& = & \lim_{x \to 0}{\frac{(2 \cos^2 x  1)  1}{\cos x  1}} \\\\
& = & \lim_{x \to 0}{\frac{2 \cos^2 x  2}{\cos x  1}} \\\\
& = & \lim_{x \to 0}{\frac{2 ( \cos^2 x  1) }{\cos x  1}}
\end{array}
}\)
Factoring the numerator, we get
\(\displaystyle{ \lim_{x \to 0}{\frac{2 (\cos x  1)(\cos x + 1) }{\cos x  1}} = \lim_{x \to 0}{\frac{2(\cos x + 1) }{1}} }\)
Direct substitution now gives us \(2(1 + 1) = 4\)
Practice B01 Final Answer 
\(\displaystyle{\lim_{x\to0}{\frac{\cos(2x)1}{\cos x1}}=4}\) 
Practice B02 
\(\displaystyle{\lim_{\theta\to1/2}{[\theta\sec(\pi\theta)]}}\) 


\(\displaystyle{\lim_{\theta\to1/2}{[\theta\sec(\pi\theta)]}}\) does not exist 
1. 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 disputable and is different from how we use the idea 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.
2. The title of this video and the comments below use the term 'infinite' to describe this limit. We use the term 'finite' to describe this limit since the xvalue is approaching a finite value.
IntegralCALC Comments:
Learn how to calculate an infinite limit. In order to determine whether or not the function has an infinite limit as x approaches a specific point, you'll need to first confirm that your function as a vertical asymptote and the point you're approaching. The function will have a vertical asymptote where it is undefined because the denominator of the function is equal to zero. Then you'll need to plug in values close to and on both sides of the number you're approaching. If the value of the function gets very large as you approach the vertical asymptote, then the limit is positive infinity. If the values get very large but negative as you approach the vertical asymptote, then the limit is negative infinity. Remember that infinite limits don't technically exist, but because it gives us so much more information about the function to say that the limit is positive of negative infinity, we'll often provide that answer instead of stating that the limit does not exist (DNE). If the onesided limits are not equal, there is no general limit.
Practice B02 Final Answer 
\(\displaystyle{\lim_{\theta\to1/2}{[\theta\sec(\pi\theta)]}}\) does not exist 
Practice B03 
\(\displaystyle{\lim_{t\to0}{\frac{\tan(6t)}{\sin(2t)}}}\) 

Practice B04 
\(\displaystyle{\lim_{\theta\to0}{\frac{1\cos(\theta)}{\theta^2}}}\) 

To save space, she dropped the limit notation in her work. Make sure you understand that you can't do that on your homework or exam.
Practice B05 
\(\displaystyle{\lim_{x\to\pi/4}{\frac{\sin(x)\cos(x)}{\cos(2x)}}}\) 

Practice B06 
\(\displaystyle{\lim_{x\to1}{\frac{\sin(x1)}{x^2+x2}}}\) 

Practice B07 
\(\displaystyle{\lim_{t\to0}{\frac{\tan(5t)}{\sin(2t)}}}\) 
