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Limits of trigonometric functions

  • Thread starter lamerali
  • Start date
  • #1
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Hello,
i'm having some trouble with evaluating limits if anyone could help me out a bit i would appreciate it. thanks in advance

evaluate the following limits:

Question 1:
lim
x -> 0 [tex]\frac{2tan^{2}x}{x^{2}}[/tex]

my answer:
u = x[tex]^{2}[/tex]
as x -> 0 u-> 0

= lim
u->0 [tex]\frac{2sin u}{u cos u}[/tex]

= 2 lim
u -> 0 [tex]\frac{sin u}{u}[/tex] (lim u-> 0) [tex]\frac{1}{cos u}[/tex]

= 2 (1) [tex]\frac{1}{cos (0)}}[/tex]
= 2 (1)(1/1)
= 2

question 2:

lim
x->0 [tex]\frac{1- cosx}{x sinx }[/tex]

my answer:
lim
x ->0 [tex]\frac{sinx}{x}[/tex] (lim x-> 0 [tex]\frac{1}{1 + cosx}[/tex]

(1)(1/2)

= 1/2

question 3:

lim
x -> 0 [tex]\frac{sin 7x}{sin 4x}[/tex]

my answer:
[tex]\frac{7sin 7x}{4 sin 4x}[/tex]

i'm not sure where to go from here on this one...i know the resulting limit will be equal to 7/4 but i dont know how to come up with this answer...i'm getting really confused

question 4:

lim
x->0 [tex]\frac{sin(cosx)}{sec x}[/tex]

= [tex]\frac{sin(cosx)}{1/cosx}[/tex]
= sin(cos) x . (cos x)
= sin(cosx[tex]^{2}[/tex])
= sin(cos (0)[tex]^{2}[/tex])
= 0.84


I am really unsure of what i am doing here...i know its a lot to look at, i really appreciate the help. Thank you!
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hello lamerali! :smile:

Question 2 is fine.

In Question 1, putting u = x2 makes tan2x tan2(x2).

You should have used L'Hôpital's rule instead (or jsut tan2 = sin2/cos2)

In Question 3, you did use L'Hôpital's rule, but you differentiated wrong … try again! :smile:

In Question 4, sin(cos x) . (cos x) = sin(cosx2) is completely wrong … and where ever did 0.84 come from? :confused:

You should just have put x = 0 … no limit is involved. :wink:
 
  • #3
62
0
Thannnnnnnk you tiny tim!!!
question 1, 2, and 3 are completely clear, i believe i came up with the correct answer. :D

However i'm still unsure of question 4. Even if you plug in x = 0 into [tex]\frac{sin cosx}{sec x}[/tex] where sec x is equal to [tex]\frac{1}{cos x}[/tex] my answer comes to 0.84147, also if you graph the function there appears to be a limit. i dont know where i am going wrong in my calculations on this one! have any ideas?

thanks again for the help! :)
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
251
However i'm still unsure of question 4. Even if you plug in x = 0 into [tex]\frac{sin cosx}{sec x}[/tex] where sec x is equal to [tex]\frac{1}{cos x}[/tex] my answer comes to 0.84147, also if you graph the function there appears to be a limit. i dont know where i am going wrong in my calculations on this one! have any ideas?
Hi lamerali! :smile:

Your technique for Question 4 seems fine now.

sin(1) = 0.84147 … you've now got the right result for the right reason! :wink:
 
  • #5
62
0
Great!! :D thank you tiny tim!!
 

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