- #1
lamerali
- 62
- 0
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 don't 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!
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 don't 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!