Jump to content
Science Forums

Calculus help. (limits)


Recommended Posts

Just reviewing a bit of calculus, and this question stumped me.


[Latex]\displaystyle\lim_{x\rightarrow2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} [/math]

Anyone got a solution? And i would appreciate an algebraic way, you dont learn anything from just typing it into your calculator.


Edit: Ive already tried rationalizing the top and bottom, neither one creates an equation in which you can derive an answer.




PS. that latex stuff can be confusing :xx:

Link to comment
Share on other sites

[math]\displaystyle\lim_{x\rightarrow2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} [/math]


This is a pretty simple question, once you figure the basics behind it.


[math]= \displaystyle\lim_{x\rightarrow2} \frac{(6-x-4)(\sqrt{3-x}+1)}{(3-x-1)(\sqrt{6-x}+2)} [/math]


See? I've rationalised it. Pretty simple now.


[math]= \displaystyle\lim_{x\rightarrow2} \frac{(2-x)(\sqrt{3-x}+1)}{(2-x)(\sqrt{6-x}+2)} [/math]


What do we see here? 2-x.


Remember that whenever an algebric term tends to zero for a value of it's variable, then you always have one or more factors that become zero at that value. All you need to do is manipulate the expression in such a way so as to 'expose' that factor.


I'd say that the things are pretty simple from here. Go on, and feel free to ask if you don't figure.

Link to comment
Share on other sites

  • 2 weeks later...

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

  • Create New...