Why does Lim as x approaches infinity of x/(x-9) = 1?

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Homework Statement



Limit as x approaches infinity of x/x-9

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The Attempt at a Solution



I know the indeterminate form infinity/infinity happens. I don't know how to fix it, but I'm assuming it's quite simple...
 
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When x is very very large, say 6 billion. The numbers 6 billion and 6 billion minus 9 are essentially the same. As x gets larger and larger, this discrepancy diminishes. The value the ratio approaches is 1.

If you wanted to be a little more rigorous, you can divide the numerator and denominator by 1/x then take the limit and see what happens. It should pop right out.
 
A nice trick for these types of problems would be to divide everything by x;

\lim_{x\rightarrow\infty} \frac{x/x}{(x/x)-(9/x)}

Then simplify and you should be able to figure it out from there
 
Using L'Hopitals rule for evaluating indeterminate forms, just differentiate both the numerator and the denominator separately to produce 1/1. Xyius explains well intuitively why the result is what it is.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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