# Limits of trigonometric functions

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 $$\frac{2tan^{2}x}{x^{2}}$$

u = x$$^{2}$$
as x -> 0 u-> 0

= lim
u->0 $$\frac{2sin u}{u cos u}$$

= 2 lim
u -> 0 $$\frac{sin u}{u}$$ (lim u-> 0) $$\frac{1}{cos u}$$

= 2 (1) $$\frac{1}{cos (0)}}$$
= 2 (1)(1/1)
= 2

question 2:

lim
x->0 $$\frac{1- cosx}{x sinx }$$

lim
x ->0 $$\frac{sinx}{x}$$ (lim x-> 0 $$\frac{1}{1 + cosx}$$

(1)(1/2)

= 1/2

question 3:

lim
x -> 0 $$\frac{sin 7x}{sin 4x}$$

$$\frac{7sin 7x}{4 sin 4x}$$

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 $$\frac{sin(cosx)}{sec x}$$

= $$\frac{sin(cosx)}{1/cosx}$$
= sin(cos) x . (cos x)
= sin(cosx$$^{2}$$)
= sin(cos (0)$$^{2}$$)
= 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!

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tiny-tim
Homework Helper
Hello lamerali! 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! In Question 4, sin(cos x) . (cos x) = sin(cosx2) is completely wrong … and where ever did 0.84 come from? You should just have put x = 0 … no limit is involved. 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 $$\frac{sin cosx}{sec x}$$ where sec x is equal to $$\frac{1}{cos x}$$ 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! :)

tiny-tim
Homework Helper
However i'm still unsure of question 4. Even if you plug in x = 0 into $$\frac{sin cosx}{sec x}$$ where sec x is equal to $$\frac{1}{cos x}$$ 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! Your technique for Question 4 seems fine now.

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