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I am working with the next limit:

$$\lim_{x \mapsto \infty }\left ( x-\sqrt{x²-2x} \right )$$

I intuitively know that since $x^2$ increases faster than $x$, when x tends to infinte this limit for a sufficient big $x$ its approximately: $$\lim_{x \mapsto \infty }\left ( x-\sqrt{x²-2x} \right )\approx \lim_{x \mapsto \infty }\left ( x-\sqrt{x²} \right )\approx0$$

when $x$ tends to infinite, Due to that fact, I think that the result of the limit it's:

$$\lim_{x \mapsto \infty }\left ( x-\sqrt{x²-2x} \right )=0$$

However, I need a some more mathematical justification rather than the intuitive justification, I would appreciate any help or hint to justify this result.

Neo
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4 Answers4

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Hint. Write, as $x \to +\infty$, $$ x-\sqrt{x^2-2x}=\frac{( x-\sqrt{x^2-2x})( x+\sqrt{x^2-2x})}{ x+\sqrt{x^2-2x}}=\frac{2x}{ x+\sqrt{x^2-2x}}\to 1 $$

Olivier Oloa
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Olivier has a fine answer with the right result.

Here is what is wrong with your intuition.

Letting $y=x-1$ then:

$$\lim_{x\to\infty} \left(x-\sqrt{x^2-2x}\right) = 1+\lim_{y\to\infty} \left(y-\sqrt{y^2-1}\right)$$

Your intuition would give that both of these limits are zero, which would give:

$$0=1+0.$$

Thomas Andrews
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A result that makes this easy is that $\sqrt{1+z} \approx 1+z/2 $ for small $z$.

More generally, $(1+z)^a \approx 1+az $ for small $z$ and real $a$.

And, of course, by Taylor, $f(x+h) \approx f(x) + hf'(x) $ for small $h$. From this, let $f(x) = (1+x)^a$.

marty cohen
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We transform the indeterminate form as follows$$x-\sqrt{x^2-2x}=x(1-\sqrt{1-2/x})=\frac {1-\sqrt{1-2/x}}{\frac1x}$$ Hence applying l'Hôpital Rule one time we get $$\frac{1}{\sqrt{1-2/x}}$$ for the limit when $x\mapsto \infty$. It is clear now that the searched limit is equal to $1$

Piquito
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