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Help with a limit

  • Thread starter sony
  • Start date
  • #1
104
0
Hi,

Lim(t->0) t / sqrt(4+t)-sqrt(4-t)

I've stared at this for like half an hour :( Could someone please give me some hints of how I start factorizing this? Thanks.
 

Answers and Replies

  • #2
MalleusScientiarum
Can you use L'Hopital's Rule?
 
  • #3
104
0
No, sorry. We havnent gotten that far yet :)
 
  • #4
TD
Homework Helper
1,022
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Multiply nominator & denominator with the complement of the denominator, i.e. [itex]\left(\sqrt {4 + t} + \sqrt {4 - t}\right)[/itex]

In the denominator, use [itex]\left( {a - b} \right)\left( {a + b} \right) = a^2 - b^2[/itex]

Now, you should be able to cancel out a t and just fill in t = 0.
 
  • #5
Curious3141
Homework Helper
2,843
87
The easiest way, use the binomial expansion (the general form for nonintegral exponents).

[tex](1 + x)^{\frac{1}{2}} \approx 1 + \frac{1}{2}x[/tex] for [itex]|x| << 1[/itex]

So

[tex]\lim_{t -> 0}\frac{t}{\sqrt{4 + t} - \sqrt{4 - t}} = \lim_{t -> 0}\frac{t}{2(\sqrt{1 + \frac{t}{4}} - \sqrt{1 - \frac{t}{4}})} = \lim_{t -> 0}\frac{t}{2[(1 + \frac{t}{8}) - (1 - \frac{t}{8})]} = \lim_{t -> 0}\frac{t}{\frac{t}{2}} = 2[/tex]
 
  • #6
104
0
TD: Then i get ( sqrt(4+t)-sqrt(4-t)) / 2

But the answer is supposed to be 2. I'm still stuck :(
 
  • #7
TD
Homework Helper
1,022
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It should be a + between those square roots in the nominator :wink:
 
  • #8
104
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Curious: I've never seen that formula before. But thanks.
 
  • #9
104
0
TD: Gaaah, stupid me. Well I got it now. Thanks!
 
  • #10
TD
Homework Helper
1,022
0
No problem :smile:

This tactic is often used to get rid off square roots.
 

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