Evaluate limit 2x-10/sqrt(x)-sqrt(5) as x -> 5

[Jade]

lim x goes to 5

2x-10/sqrt(x)-sqrt(5)

When you plug in 5 it becomes an indeterminate case 0/0

So then you must multiply by the conjugate sqrt(x)+sqrt(5)

I am having trouble multiplying the top

2x-10(sqrt(x)+sqrt(5)/x-5

 

[pka]

2x-10=2(x-5), now divide!

 

[Jade]

Ah - ha!!!!

2(sqrt(x)+sqrt(5))

 

[pka]

The final answer is \(\displaystyle \L
4\sqrt 5 .\)

 

[devan]

Yeah @ pka.. what you do is use l'hopital's rule [if you've hear of it]. That is, you differentiate the numerator and the denominator to get:

[BY L'HOPITAL'S]

lim 2x-10/sqrt[x]-sqrt[5] = lim 2/.5[sqrt[x]] = lim (2x2)sqrt[x] = 4[sqrt[5]]
x->5 x-> 5 x->5

so thats about it[/tex]

 

[skeeter]

Yeah @ pka.. what you do is use l'hopital's rule [if you've hear of it].

I'm quite sure he's heard of it.

Be advised that L'Hopital's rule is not taught or used at this level of introductory calculus.