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You CAN Ace Calculus

17calculus > limits > trig limits

### Calculus Main Topics

Limits

Single Variable Calculus

Multi-Variable Calculus

### Tools

math tools

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Limits Involving Trig and Inverse 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.

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.

### Search 17Calculus

Practice Problems

Instructions - Unless otherwise instructed, evaluate these limits. Give your answers in exact terms.

 Level A - Basic

Practice A01

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

solution

Practice A02

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

solution

Practice A03

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

solution

Practice A04

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

solution

Practice A05

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

solution

Practice A06

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

solution

Practice A07

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

solution

Practice A08

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

solution

Practice A09

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

solution

Practice A10

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

solution

Practice A11

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

solution

Practice A12

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

solution

Practice A13

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

solution

Practice A14

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

solution

Practice A15

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

solution

Practice A16

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

solution

 Level B - Intermediate

Practice B01

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

solution

Practice B02

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

solution

Practice B03

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

solution

Practice B04

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

solution

Practice B05

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

solution

Practice B06

$$\displaystyle{\lim_{x\to1}{\frac{\sin(x-1)}{x^2+x-2}}}$$

solution

Practice B07

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

solution

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