- #1

- 104

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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.

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- Thread starter sony
- Start date

- #1

- 104

- 0

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.

- #2

MalleusScientiarum

Can you use L'Hopital's Rule?

- #3

- 104

- 0

No, sorry. We havnent gotten that far yet :)

- #4

TD

Homework Helper

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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

- #5

Curious3141

Homework Helper

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[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

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TD: Then i get ( sqrt(4+t)-sqrt(4-t)) / 2

But the answer is supposed to be 2. I'm still stuck :(

But the answer is supposed to be 2. I'm still stuck :(

- #7

TD

Homework Helper

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It should be a **+** between those square roots in the nominator

- #8

- 104

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Curious: I've never seen that formula before. But thanks.

- #9

- 104

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TD: Gaaah, stupid me. Well I got it now. Thanks!

- #10

TD

Homework Helper

- 1,022

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No problem

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

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

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