What is the limit as x approaches -inf of sqrt(x^2+5x+1) -x

I've tried to do this using the conjugate and I got -5 but my answer sheer tells me thats not it...but I don't know how to get the right answer

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- Oct 6th 2009, 08:54 PMkatekatelimit as x approaches -infinity?
What is the limit as x approaches -inf of sqrt(x^2+5x+1) -x

I've tried to do this using the conjugate and I got -5 but my answer sheer tells me thats not it...but I don't know how to get the right answer - Oct 6th 2009, 09:20 PMCaptainBlack
- Oct 6th 2009, 09:27 PMtonio

Of course the answer cannot negative: square root is alwasy positive, and the -x part is ALSO positive sicne x --> -oo ! ==> the limit, in case it exists and is finite, must be non-negative.

Now, the conjugate thing is a good idea, and then we get:

[x^2 + 5x + 1 - x^2]/[Sqrt(x^2+5x+1) + x] = [5x + 1]/[Sqrt(x^2+5x+1) + x]

The trick now is to mutiply by (1/-x)/(1/-x)....why -1/x? Because we'll want to "introduce" 1/-x into the square, but as x --> -oo we get that for almost all the values of x it is negative and we need something POSITIVE inside a square root, so:

{[5x + 1]/[Sqrt(x^2+5x+1) + x]}(1/-x)/(1/-x) = [-5 - 1/x]/[Sqrt(1 + 5/x + 1/x^2) - 1] --- pay attention to the fact that we're introducing 1/-x into the Square root BUT already inside it becomes 1/x^2, of course.

We can now see that the numerator --> -5 whereas the denominator --> 0 (from the left since 1 + 5/x + 1/x^2 < 1...remember: x < 0) ==> the limit doesn't exist in finite form, and in fact the expression --> oo as x --> -oo

Tonio - Oct 6th 2009, 09:27 PMkatekate
That's definitely the right expression, ending with -x and to -inf

- Oct 6th 2009, 09:28 PMmmattson07
try substituting really large negative numbers for x...the result are really large pos numbers

- Oct 7th 2009, 12:48 AMCaptainBlack