evaluate limit

[bt359]

hello, pls help me or give me some hints how to solve this... I have try to multiply everything by its conjugates but it didnt work... I dont know what i am doing wrong. the awser should be (3/2)Thanks in advance

.\(\displaystyle \displaystyle{\lim_{\x\, \rightarrow\, 2}\, \frac{sqrt(x^2+4)\,-\,2}{sqrt(x^2+9)\,-\,3}\, }\)

 

[Deleted member 4993]

hello, pls help me or give me some hints how to solve this... I have try to multiply everything by its conjugates but it didnt work... I dont know what i am doing wrong. the awser should be (3/2)Thanks in advance

.\(\displaystyle \displaystyle{\lim_{\ x\, \rightarrow\, 2}\, \frac{\sqrt{x^2+4}\,-\,2}{\sqrt{x^2+9}\,-\,3}\, }\)

.

I edited your problem above. Is that the correct problem?

If it is - I don't see how the answer could be 3/2!

Please check and re-post

If the problem is .\(\displaystyle \displaystyle{\lim_{\ x\, \rightarrow\ 0}}\) - then the answer would be 3/2.

In that case, apply L'Hospital's Rule.

 

[pka]

hello, pls help me or give me some hints how to solve this... I have try to multiply everything by its conjugates but it didnt work... I dont know what i am doing wrong. the awser should be (3/2)Thanks in advance
​\(\displaystyle \displaystyle{\lim_{x\rightarrow 2}\, \frac{\sqrt{x^2+4}\,-\,2}{\sqrt{x^2+9}\,-\,3}\, }\)

=\(\displaystyle \displaystyle{\frac{\sqrt{8}\,-\,2}{\sqrt{13}\,-\,3}\, }\)

 

[HallsofIvy]

hello, pls help me or give me some hints how to solve this... I have try to multiply everything by its conjugates but it didnt work... I dont know what i am doing wrong. the awser should be (3/2)Thanks in advance

.\(\displaystyle \displaystyle{\lim_{\x\, \rightarrow\, 2}\, \frac{sqrt(x^2+4)\,-\,2}{sqrt(x^2+9)\,-\,3}\, }\)

Here's the hint- go back and read the problem again! I supspect you will find it is the limit as x goes to 0, not 2.

(If it does say "\(\displaystyle x\to 2\)", it is wrong. Take the limit as x goes to 0.)

 

[bt359]

Limits.

closed