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Example
lim
ℎ→
𝜋
4
8𝑥 tan 𝑥 − 2𝜋 tan 𝑥
4𝑥 − 𝜋
= lim
ℎ→
𝜋
4
2 tan 𝑥 [ 4𝑥−𝜋]
4𝑥− 𝜋
lim
ℎ→
𝜋
4
2 tan 𝑥 [ 4𝑥−𝜋]
4𝑥− 𝜋
( 𝑎𝑝𝑝𝑙𝑦𝑖𝑛𝑔𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑛𝑔 𝑐𝑜𝑚𝑚𝑜𝑛 𝑚𝑜𝑛𝑜𝑚𝑖𝑎𝑙 𝑓𝑎𝑐𝑡𝑜𝑟)
= lim
ℎ→
𝜋
4
2 tan 𝑥
= 2tan
𝜋
4
= 2 (1)](https://image.slidesharecdn.com/limits-involving-trigonometric-functions-241123125902-87a6a031/75/Limits-Involving-Trigonometric-Functions-pdf-12-2048.jpg)
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- 1. SLIDESMANIA.COM Title Here Title Here TitleHere Title Here Title Here Title Here Title Here Title Here Limits Involving Trigonometric Functions
- 2. SLIDESMANIA.COM Learning Objectives: Find thelimits involving trigonometric functions using appropriate theorems.
- 3. SLIDESMANIA.COM Title Here TitleHere Title Here Title Here Title Here Title Here Review REview
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- 5. SLIDESMANIA.COM The techniques andrules for finding limits discussed in the previous lessons apply to the trigonometric and exponential functions as well as to algebraic functions. 1. lim 𝑥→0 sin 𝑥 𝑥 = 𝑥 𝑠𝑖𝑛 𝑥 = 1 2. lim 𝑥→0 1−cos 𝑥 𝑥 = 0
- 6. SLIDESMANIA.COM Title Here TitleHere Title Here Title Here Title Here Title Here 1.Find the lim 𝑥→ 𝜋 4 sin 𝑥 𝑐𝑜𝑠𝑥 Example Title Here lim 𝑥→ 𝜋 4 sin 𝑥 𝑐𝑜𝑠𝑥 = lim 𝑥→ 𝜋 4 sin 𝜋 4 cos 𝜋 4 = 2 2 ∙ 2 2 = 1
- 7. SLIDESMANIA.COM Example lim 𝑥→0 1 − cos𝑥 𝑠𝑖𝑛 𝑥 ∙ 1 + cos 𝑥 1 + cos 𝑥 = 1−cos2 𝑥 (sin 𝑥) ( 1+cos 𝑥) = sin2 𝑥 (sin 𝑥) ( 1+cos 𝑥) = (sin 𝑥) (sin 𝑥) (sin 𝑥) ( 1+cos 𝑥) = (sin 𝑥) (1+cos 𝑥) Therefore, lim 𝑥→0 1−cos 𝑥 𝑠𝑖𝑛 𝑥 = = (sin 𝑥) (1+cos 𝑥) = 0 1+1 = 0 (rationalize) Find the lim 𝑥→0 1 − cos 𝑥 𝑠𝑖𝑛 𝑥
- 8. SLIDESMANIA.COM Title Here Title Here TitleHere Title Here Title Here = lim 𝑥→0 sin 3𝑥 𝑥 = 3 lim 𝑥→0 sin 3𝑥 3𝑥 = 3 ( 1) = 3 Example Title Here Find the lim 𝑥→0 sin 3𝑥 𝑥 Note that : lim 𝑥→0 sin 𝑥 𝑥 = 𝑥 𝑠𝑖𝑛 𝑥 = 1 We can factor out the 3 from inside the sine function to make the argument of the sine function match the standard form. Example
- 9. SLIDESMANIA.COM Title Here Title Here TitleHere Example Title Here Find the lim 𝑥→0 3𝑥 tan 𝑥 sin 𝑥 = lim 𝑥→0 3 0 cos 0 = 0 = lim 𝑥→0 3𝑥 sin 𝑥 cos 𝑥 sin 𝑥
- 10. SLIDESMANIA.COM 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 lim 𝑥→𝜋 cos𝑥 𝑥2 = lim 𝑥→𝜋 cos 𝜋 𝑥2 = −1 𝜋2 Example Note : 𝜋 = 180 in degree
- 11. SLIDESMANIA.COM Example lim ℎ→3 sin( ℎ −3) ℎ2 + 2ℎ − 15 = lim ℎ→3 sin( ℎ−3) ℎ+5 ( ℎ−3) ( 𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑛𝑔) = lim ℎ→3 1 ℎ+5 ∙ sin( ℎ−3) ℎ−3 = 1 3+5 ∙ 1 = 1 8 NOTE:
- 12. SLIDESMANIA.COM Example lim ℎ→ 𝜋 4 8𝑥 tan 𝑥− 2𝜋 tan 𝑥 4𝑥 − 𝜋 = lim ℎ→ 𝜋 4 2 tan 𝑥 [ 4𝑥−𝜋] 4𝑥− 𝜋 lim ℎ→ 𝜋 4 2 tan 𝑥 [ 4𝑥−𝜋] 4𝑥− 𝜋 ( 𝑎𝑝𝑝𝑙𝑦𝑖𝑛𝑔𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑛𝑔 𝑐𝑜𝑚𝑚𝑜𝑛 𝑚𝑜𝑛𝑜𝑚𝑖𝑎𝑙 𝑓𝑎𝑐𝑡𝑜𝑟) = lim ℎ→ 𝜋 4 2 tan 𝑥 = 2tan 𝜋 4 = 2 (1)
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- 19. SLIDESMANIA.COM Title Here Title Here 1.lim 𝑥→ 𝜋 2 cot 𝑥 2. lim 𝑥→ 𝜋 4 3 tan 𝑥 + 1 3. lim 𝑥→0 sin 5𝑥 𝑥 4. lim 𝑥→0 tan 𝑥 4𝑥 5. lim 𝑥→0 sin 5𝑥 sin 6𝑥 Determine the value of each limits if it exist. 6. lim 𝑥→1 sin( 𝑥2−1) 𝑥−1 7. lim 𝑥→11 cos( 𝑥−7) 𝑥2−16𝑥+63 • 1 5𝑥 8. lim 𝑥→0 cos2 𝑥 −cos(𝑥) cos(𝑥)−1 9. lim 𝑥→2 sin( 2𝑥−4) 5𝑥 −10
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