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Limit to negative infinity

  • Thread starter dnt
  • Start date
dnt
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1. Homework Statement

limit as x-> -oo of x+(x^2+3)^(1/2) (square root)

2. Homework Equations

n/a

3. The Attempt at a Solution

i first multiplied the top and bottom (which is just 1) by the conjugate to get:

-3
--------------
[x - (x^2+3)^(1/2)]

then i divided by x on top and bottom to get:

-3/x
-----------
[x/x - (1 + 3/x2)^(1/2)]

but now what do i do? the top goes to 0 but the bottom also goes to 0 (because its 1 - (1+0)^(1/2)). but the answer is suppose to be 0.

0/0 isnt 0. can someone point me in the right direction? thanks.
 

Answers and Replies

dextercioby
Science Advisor
Homework Helper
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[tex] \lim_{x\rightarrow -\infty} x+\sqrt{x^{2}+3} =\lim_{x\rightarrow \infty}
\sqrt{x^{2}+3}-x [/tex]

After multiplying by the conjugate expression you're facing the limit

[tex] \lim_{x\rightarrow \infty} \frac{3}{\sqrt{x^{2}+3}+x} [/tex]

which is trivial.

Daniel.
 

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