- #1

- 13

- 0

lim (x-1)^1/2 - 2/ x^2-25

x 5

lim x + 1-e^x/ x^3

x 0+

lim x^3/2 + 5x - 4/ x ln x

x infinity

lim tan x ln sin x

x pie/2-

lim (cos x)^x+1

x 0

lim (1+1/x)^5x

x infinity

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Tonya Miller
- Start date

- #1

- 13

- 0

lim (x-1)^1/2 - 2/ x^2-25

x 5

lim x + 1-e^x/ x^3

x 0+

lim x^3/2 + 5x - 4/ x ln x

x infinity

lim tan x ln sin x

x pie/2-

lim (cos x)^x+1

x 0

lim (1+1/x)^5x

x infinity

- #2

- 221

- 0

I assume you mean...Tonya Miller said:Explain how to complete these problems:

lim (x-1)^1/2 - 2/ x^2-25

x 5

[tex]

\lim_{x -> 5} \frac{\sqrt(x-1) -2}{x^2-25}

[/tex]

To do this problem just multiply the top and bottom by the conjugate of the numerator....use the difference of square to factor the x^2-25 and you will have a common factor to take out....

lim x + 1-e^x/ x^3

x 0+

On this one I really can't say without knowing exactly what is in the numerator and denominator... try using parentheses so I know how things are grouped.

I need to know what you mean here too.lim x^3/2 + 5x - 4/ x ln x

x infinity

lim tan x ln sin x

x pie/2-

I haven't brushed up on indeterminate forms but I believe this is a case where you can apply L.H....but not directly...write as [tex]ln(sin(x))/cot(x)[/tex] and then you have a 0/0....L.H. gives you...check my work...-cos(x)*sin(x) And this limit is clearly 0.

Why not just plug in 0 directly? Looks like 2 to me.lim (cos x)^x+1

x 0

lim (1+1/x)^5x

x infinity

Well if you wrote it like

[tex]

((1+ \frac{1}{x})^x)^5

[/tex]

And you know what the inside limit is....so did euler....then just raise that to the fifth power.

- #3

- 176

- 0

hello there

do you want to prove the limit exists? or do you want to find the limit?

steven

do you want to prove the limit exists? or do you want to find the limit?

steven

- #4

- 13

- 0

#2 lim (x+1-e^x)/(x^3)

x=>0+

#3 lim (x^3/2 +5x-4)/(xlnx)

x=>infinity

- #5

- 13

- 0

infinity

S x/x^4+9 dx

neg.infinity

0

S 1/(x-8)^2/3 dx

neg.infinity

0

S x/(4-x^2)1/2 dx

-2

pie

S 1/(1-cos x) dx

0

- #6

- 13

- 0

infinity

E n/(n^2 + 1) x^n

n=2

E [(1)(3)(5)(7)...(2n-1)]/[(3)(6)(9)(12)...(3n)] (x-2)^n

- #7

- 13

- 0

1) {n/n+1}

2) E 2^n/(n^2)

3) E (n!)^2/ (2n)!

4) E (n 3^2n)/ (5^n-1)

5) E (-1)^n+1 n/(n^2 +4)

infinity

6) E (-1)^n n/(ln n)

n= 2

- #8

- 13

- 0

1) x=t^3, y= t^2

2) x = sec t + 2, y = tan t -1

- #9

- 13

- 0

1) x = (t)^1/2 , y = 3t + 4, t = 4

- #10

- 13

- 0

1) S 1/ (x)^1/2 [(x)^1/2 + 1] dx

2) S x atn x^2 dx

3) S x 4^-x2 dx

4) S x^2/ x+ 2 dx

5) S (e^2x + e^3x)^2/ (e^5x) dx

6) S csc(1+ cot(x)) dx

- #11

- 13

- 0

1) S (x^2+3x+5) e^3x dx

2) S e^4x sin(5x) dx

3) S sec^3 (3x) dx

4) S sin^2 4x cos^2 4x dx

5) sec x/ (cot^5 x) dx

6) (4-x^2)^1/2 / (x^2) dx

- #12

- 13

- 0

1) S 3x-5 / (1 + x^2)^1/2 dx

2) S 5x^3- 3x^2 + 7x - 3/ (x^4 + 2x^2 + 1) dx

3) S 5x^2 + 11x +17/ (x^3 + 5x^2 + 4x + 20) dx

- #13

- 13

- 0

1) y = ln (x/ 3x + 5))^4

2) y = (ln (x)^1/2)^1/2

3) y = 5^3x + (3x)^5

4) y = x^5x+1

5) y = (sin x )^ cos x

- #14

- 13

- 0

#2 lim (x+1-e^x)/(x^3)

x=>0+

#3 lim (x^3/2 +5x-4)/(xlnx)

x=>infinity

Could you show how to do these integrals?

infinity

S x/x^4+9 dx

neg.infinity

0

S 1/(x-8)^2/3 dx

neg.infinity

0

S x/(4-x^2)1/2 dx

-2

pie

S 1/(1-cos x) dx

0

- #15

- 13

- 0

infinity

E n/(n^2 + 1) x^n

n=2

E [(1)(3)(5)(7)...(2n-1)]/[(3)(6)(9)(12)...(3n)] (x-2)^n

Could you show in details how to converge or diverge these problems?

1) {n/n+1}

2) E 2^n/(n^2)

3) E (n!)^2/ (2n)!

4) E (n 3^2n)/ (5^n-1)

5) E (-1)^n+1 n/(n^2 +4)

infinity

6) E (-1)^n n/(ln n)

n= 2

- #16

- 221

- 0

https://www.physicsforums.com/showthread.php?t=44101

If you want to get help....at the very least write down what you think the solution to each one is....

- #17

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

Share: