Limits

[aniel]

the first one: sqrt(x + 3) - 2 / 4x -4
the second one : 1 - sqrt(cosx) / x^2
thnk you

 

[Jameson]

Hi aniel,

Welcome to Math Help Boards!

Is the first limit this? \(\displaystyle \frac{\sqrt{x+3}-2}{4x-4}\)?

Is the second one this? \(\displaystyle \frac{1-\sqrt{\cos{x}}}{x^2}\)?

The other thing is limits must have x approach some value, so we can't do anything with these two problems until we know that. You'll usually see something like \(\displaystyle \lim_{x \rightarrow 3}\).

 

[aniel]

Im sorry i forgot where the x tends to go :/
lim (x-> 1) sqrt (x+3) - 2 /4x - 4
lim (x->0) 1- sqrt (cosx) / x^2

and im so sorry taking your time but i get so confused with limits .

 

[Jameson]

Im sorry i forgot where the x tends to go :/
lim (x-> 1) sqrt (x+3) - 2 /4x - 4
lim (x->0) 1- sqrt (cosx) / x^2

and im so sorry taking your time but i get so confused with limits .

Don't be sorry at all This site was created so anyone can get free math help.

Ok, next question. Have you covered L'Hôpital's rule in your calculus class? Both limits end up being 0/0 and that is a special case where we can use a time saving shortcut.

 

[aniel]

Welll thank you sir
Yes we used the hospital rule and well i did some shortcuts at both of them but the book solution it doesnt coincides with my solution.

well i went till here to both of them
lim (x-> 1) sqrt (x+3) - 2 /4x - 4 = §
lim (x->1) [sqrt( (x+3) -2] * [sqrt( (x+3) +2]/ (4x-4) *sqrt( (x+3) +2 =
lim (x->1) x+3-4 /4x-4 [sqrt( (x+3) +2]

and to the second one:
ilm (x->0) 1- sqrt (cosx) / x^2
=lim (x->0) [1- sqrt (cosx)]*[1+ sqrt (cosx)] / x^2 * [sqrt(cosx)
=lim (x->0) sin^2x /[1 + sqrt(cosx)] *x^2

 

[Jameson]

Welll thank you sir
Yes we used the hospital rule and well i did some shortcuts at both of them but the book solution it doesnt coincides with my solution.

You're doing them "the long way", which is fine just longer. I'll do the first one the way you started it though to get the answer

\(\displaystyle \lim_{x \rightarrow 1} \frac{\sqrt{x+3}-2}{4x-4} = \frac{\sqrt{x+3}-2}{4x-4} \cdot \frac{\sqrt{x+3}+2}{\sqrt{x+3}+2}=\frac{x-1}{(4x-4)(\sqrt{x+3}+2)}\)

That's pretty much what you got to and is correct thus far. The "trick" here is seeing that \(\displaystyle 4x-4=4(x-1)\). This will allow us to make a nice cancellation.

\(\displaystyle \frac{x-1}{(4x-4)(\sqrt{x+3}+2)}=\frac{x-1}{4(x-1)(\sqrt{x+3}+2)}=\frac{1}{(4)(\sqrt{x+3}+2)}\)

I haven't checked this solution, but I'm hoping that it's 1/16, which is what you get if you now direct substitute.

 

[aniel]

Than you sir yes i solved it and it is 1/16
my mistake its that i did in this way 4(x - 4/x) :/

 

[MarkFL]

I get the same result using L'Hôpital's rule:

$\displaystyle \lim_{x\to1}\frac{\sqrt{x+3}-2}{4x-4}=\lim_{x\to1}\frac{\frac{1}{2\sqrt{x+3}}}{4}= \frac{1}{16}$

 

[Jameson]

Than you sir yes i solved it and it is 1/16
my mistake its that i did in this way 4(x - 4/x) :/

It happens. We all make mistakes like this. Just have to learn from them. You are doing really well though because you knew that you would have to cancel something in order to solve the problem, which is the key idea with these "indeterminate forms". Someone else is replying to this thread now, so I'll won't comment on #2 yet.

Once again, welcome to MHB! I hope you stick around and participate in our community.

 

[aniel]

Well im really thankfull sir...its my pleasure to be part of this group, i hope i can learn more now and on.
Congratulations about ur job here.Its a great thing .
Thank you again !

 

[MarkFL]

Have you had any success with the second limit? Applying L'Hôpital's rule twice works nicely.

 

[soroban]

Hello, aniel!

Here's the first one.

[tex]\lim_{x\to1}\frac{\sqrt{x + 3} - 2}{4x -4 }[/tex]

Rationalize the numerator:

[tex]\frac{\sqrt{x+3} - 2}{4x-4}\cdot\frac{\sqrt{x+3}+2}{\sqrt{x+3}+2} \;=\; \frac{(x+3) - 4}{(4x-4)(\sqrt{x+3}+2)} [/tex]

. . .[tex]=\;\frac{x-1}{4(x-1)(\sqrt{x+3}+2)} \;=\;\frac{1}{4(\sqrt{x+3}+2)} [/tex]


Therefore: .[tex]\lim_{x\to1}\frac{1}{4(\sqrt{x+3}+2)} \;=\;\frac{1}{4(\sqrt{4}+2)} \;=\;\frac{1}{4(2+2)} \;=\;\frac{1}{16}[/tex]