lim x-->9 (x sqrt(3) - 3) / (x - 9): please help checking , must be right!!!!!!

[nogoodmaths]

Please point out what is wrong and help me correct them.

Evaluate the following limit, if exits

1.lim x-->9 (x sqrt(3) - 3) / (x - 9)

my answertwo-sided limit does not exist)
lim_(x->9^-) (x sqrt(3)-3)/(x-9) = -infinity (Limit from the left)
lim_(x->9^+) (x sqrt(3)-3)/(x-9) = infinity (limit from the right)

2a)lim x-->∞ (3 + 2/x) * cos(1/x)

Find the following limit:
lim_(x->infinity) cos(1/x) (3+2/x)
Applying the product rule, write lim_(x->infinity) (3+2/x) cos(1/x) as (lim_(x->infinity) (3+2/x)) (lim_(x->infinity) cos(1/x)):
lim_(x->infinity) (3+2/x) lim_(x->infinity) cos(1/x)
Using the fact that cosine is a continuous function, write lim_(x->infinity) cos(1/x) as cos(lim_(x->infinity) 1/x):
lim_(x->infinity) (3+2/x) cos(lim_(x->infinity) 1/x)
Let epsilon>0. Then for all x > 1/epsilon, 1/x<1/(1/epsilon) = epsilon, so
lim_(x->infinity) 1/x = 0:
cos(0) lim_(x->infinity) (3+2/x)
cos(0) = 1:
lim_(x->infinity) (3+2/x)
3+2/x = lim_(x->infinity) 3+2 (lim_(x->infinity) 1/x):
lim_(x->infinity) 3+2 lim_(x->infinity) 1/x
Let epsilon>0. Then for all x > 1/epsilon, 1/x<1/(1/epsilon) = epsilon, so
lim_(x->infinity) 1/x = 0:
lim_(x->infinity) 3+2×0
Since 3 is constant, lim_(x->infinity) 3 = 3:
3+2×0
3+2 0 = 3:
Answer: |=3

2b.)lim x-->0 ((1 - cos(ax)) / x^2)

Find the following limit:
lim_(x->0) (1-cos(a x))/x^2
Applying l'Hôpital's rule, we get that
lim_(x->0) (1-cos(a x))/x^2
= lim_(x->0) ( d/( dx)(1-cos(a x)))/( d/( dx) x^2)
= lim_(x->0) (a sin(a x))/(2 x)
lim_(x->0) (a sin(a x))/(2 x)
(a lim_(x->0) (sin(a x))/x)/(2)
Applying l'Hôpital's rule, we get that
lim_(x->0) (sin(a x))/x
| = | lim_(x->0) ( d/( dx) sin(a x))/(( dx)/( dx))
| = | lim_(x->0) (a cos(a x))/1
| = | lim_(x->0) a cos(a x)
(a lim_(x->0) a cos(a x))/2
lim_(x->0) a cos(a x) = a cos(0 a) = a: (a a)/2 1/2 a a = a^2/2:

Answer: | | a^2/2

3.)lim x-->∞ (x^1/3 - 5x + 3) / (2x + x^2/3 - 4)

Find the following limit:
lim_(x->infinity) (3+x^(1/3)-5 x)/(x^2/3+2 x-4)
(3+x^(1/3)-5 x)/(x^2/3+2 x-4) = (3+x^(1/3)-5 x)/(x^2/3+2 x-4):
lim_(x->infinity) (3+x^(1/3)-5 x)/(x^2/3+2 x-4)
The leading term in the denominator of (3+x^(1/3)-5 x)/(x^2/3+2 x-4) is x^2.
Divide the numerator and denominator by this:
lim_(x->infinity) (3/x^2+1/x^(5/3)-5/x)/(1/3+2/x-4/x^2)
The expressions 3/x^2, 1/x^(5/3), -5/x, -4/x^2 and 2/x all tend to zero as x approaches infinity:
0/(1/3) 0/(1/3) = 0:

Answer: | | 0

4.lim x-->∞ (sqrt(x^2 + x)) - sqrt(x^2 - cx), c=positive constant

the limit:lim_(x->infinity) (sqrt(x^2+x)-sqrt(x^2-c x)) = (1+c)/2
this quaestion is the one I don't know how to deal with (no.4), please help!!!

5.lim x-->-∞ (x^1/3 - 5x + 3) / (2x + x^2/3 - 4)
divide each term by x

lim x-->-∞ ( x^1/3/x - 5x/x + 3/x) / (2x/x + x^2/3/x -4/x)

[ 0 -5 + 0] / [2 + 0 -4] =

-5/2

 

[ksdhart]

I'd take a look at your problems and make sure you copied them down correctly. I noticed that your problems #3 and #5 are the same limit, but you get two different answers (the answer for #5 is the correct one by the way). That said, here's my first few steps for #4 and hopefully that opens the way for you to proceed:

Start with the given limit:

\(\displaystyle \lim _{x\to \infty }\left(\sqrt{x^2+x}-\sqrt{x^2-cx}\right)\)

Use this algebraic property:

\(\displaystyle a-b=\left(a-b\right)\cdot \frac{a+b}{a+b}=\frac{\left(a-b\right)\left(a+b\right)}{a+b}=\frac{a^2-b^2}{a+b}\)

Let \(\displaystyle a=\sqrt{x^2+x}\) and \(\displaystyle b=\sqrt{x^2-cx}\) and simplify a bit to get this limit:

\(\displaystyle \lim _{x\to \infty }\left(\frac{\left(\sqrt{x^2+x}\right)^2-\left(\sqrt{x^2-cx}\right)^2}{\sqrt{x^2+x}+\sqrt{x^2-cx}}\right)\)

Clean up the numerator:

\(\displaystyle \lim _{x\to \infty }\left(\frac{\left(c+1\right)\cdot x}{\sqrt{x^2+x}+\sqrt{x^2-cx}}\right)\)

Now you try continuing from here. If you get stuck again, that's okay. But when you reply back, please include all of your workings (even if you know for sure they're wrong). Thank you.

 

[nogoodmaths]

re:ksdhart

I'd take a look at your problems and make sure you copied them down correctly. I noticed that your problems #3 and #5 are the same limit, but you get two different answers (the answer for #5 is the correct one by the way). That said, here's my first few steps for #4 and hopefully that opens the way for you to proceed:

Start with the given limit:

\(\displaystyle \lim _{x\to \infty }\left(\sqrt{x^2+x}-\sqrt{x^2-cx}\right)\)

Use this algebraic property:

\(\displaystyle a-b=\left(a-b\right)\cdot \frac{a+b}{a+b}=\frac{\left(a-b\right)\left(a+b\right)}{a+b}=\frac{a^2-b^2}{a+b}\)

Let \(\displaystyle a=\sqrt{x^2+x}\) and \(\displaystyle b=\sqrt{x^2-cx}\) and simplify a bit to get this limit:

\(\displaystyle \lim _{x\to \infty }\left(\frac{\left(\sqrt{x^2+x}\right)^2-\left(\sqrt{x^2-cx}\right)^2}{\sqrt{x^2+x}+\sqrt{x^2-cx}}\right)\)

Clean up the numerator:

\(\displaystyle \lim _{x\to \infty }\left(\frac{\left(c+1\right)\cdot x}{\sqrt{x^2+x}+\sqrt{x^2-cx}}\right)\)

Now you try continuing from here. If you get stuck again, that's okay. But when you reply back, please include all of your workings (even if you know for sure they're wrong). Thank you.

Here is my working, am I right? Please show me steps if wrong.