- #1

SprucerMoose

- 62

- 0

Gday,

I was contemplating limits today and tried to prove something to myself and I can't figure it out.

This is what I am trying to wrap my head around.

http://latex.codecogs.com/gif.latex?\lim_{x \to \infty } \sqrt{x^2 + 1} [Broken]

I'm not looking for the "numerical" solution of infinity, what I am trying to prove is that as x approaches infinity, this expression approaches x.

I then came to this conclusion.

http://latex.codecogs.com/gif.latex?\lim_{x \to \infty } \sqrt{x^2 + 1} = \lim_{x \to \infty } \sqrt{x^2(1 + \frac{1}{x^2})} = \lim_{x \to \infty } x\sqrt{1 + \frac{1}{x^2}} = x\sqrt{1 + 0} = x [Broken]

Is this logic correct? I can't see why it wouldn't be. But this way of thinking leads to a scenario that I am not happy with.

http://latex.codecogs.com/gif.latex?\lim_{x \to \infty } \sqrt{(x-1)^2 + 1} = \lim_{x \to \infty } \sqrt{x^2 - 2x +2} = \lim_{x \to \infty } \sqrt{x^2( 1 - \frac{2}{x} +\frac{2}{x^2})} = \lim_{x \to \infty} x\sqrt{1 - \frac{2}{x} + \frac{2}{x^2}} = x\sqrt{1 - 0 + 0} = x [Broken]

I am reasonably certain that the solution to this particular expression should be http://latex.codecogs.com/gif.latex?\lim_{x \to \infty } \sqrt{(x-1)^2 + 1} = x - 1 [Broken].

Can someone please tell me where I have gone wrong?

I was contemplating limits today and tried to prove something to myself and I can't figure it out.

This is what I am trying to wrap my head around.

http://latex.codecogs.com/gif.latex?\lim_{x \to \infty } \sqrt{x^2 + 1} [Broken]

I'm not looking for the "numerical" solution of infinity, what I am trying to prove is that as x approaches infinity, this expression approaches x.

I then came to this conclusion.

http://latex.codecogs.com/gif.latex?\lim_{x \to \infty } \sqrt{x^2 + 1} = \lim_{x \to \infty } \sqrt{x^2(1 + \frac{1}{x^2})} = \lim_{x \to \infty } x\sqrt{1 + \frac{1}{x^2}} = x\sqrt{1 + 0} = x [Broken]

Is this logic correct? I can't see why it wouldn't be. But this way of thinking leads to a scenario that I am not happy with.

http://latex.codecogs.com/gif.latex?\lim_{x \to \infty } \sqrt{(x-1)^2 + 1} = \lim_{x \to \infty } \sqrt{x^2 - 2x +2} = \lim_{x \to \infty } \sqrt{x^2( 1 - \frac{2}{x} +\frac{2}{x^2})} = \lim_{x \to \infty} x\sqrt{1 - \frac{2}{x} + \frac{2}{x^2}} = x\sqrt{1 - 0 + 0} = x [Broken]

I am reasonably certain that the solution to this particular expression should be http://latex.codecogs.com/gif.latex?\lim_{x \to \infty } \sqrt{(x-1)^2 + 1} = x - 1 [Broken].

Can someone please tell me where I have gone wrong?

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