Limits of trigonometric functions

[lamerali]

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}}

my answer:
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 }

my answer:
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}

my answer:
\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 don't 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!

 

[tiny-tim]

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 just 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.

 

[lamerali]

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 don't know where i am going wrong in my calculations on this one! have any ideas?

thanks again for the help! :)

 

[tiny-tim]

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 don't 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!

 

[lamerali]

Great! :D thank you tiny tim!