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Using the $\epsilon−\delta$ definition of limits, prove:

$$\lim\limits_{x \to 1} \frac{x^5+1}{x}=2$$

The factor $(x-1)$ I can control. And I can also limit the other factor in the numerator.

But the $x$ in the denominator is my problem because if I limit $(x-1)$ it seems to grow. I'm not sure what to do with it.

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  • $\begingroup$ I'm not sure if this could be useful to you but note that $x^5-2x+1 =(x-1)(x^4+x^3+x^2+x-1)$ $\endgroup$ – rapidracim Mar 7 '18 at 22:04
  • $\begingroup$ I would suggest doing the two limit proofs separately (and then use $\epsilon/2$ to put them together). Do $\lim x^4$ and $\lim 1/x$. If you've never done one with a denominator, you'll have to rethink your inequalities a little bit. $\endgroup$ – Ted Shifrin Mar 7 '18 at 22:05
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hint

$$x^5+1-2x=$$ $$(x-1)(x^4+x^3+x^2+x-1) $$

As $x $ is near $1$, we can assume that $$|x-1|<\frac {1}{2} $$ or $$\frac {1}{2}<x <\frac {3}{2} $$

Find $M >0$ such that $$|\frac {x^5-2x+1}{x}|<M|x-1|$$

then you take $$\delta=\min (\frac {1}{2},\frac {\epsilon}{M} )$$

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due to the linearity of the limit, it is equivalent to show that $\lim_{x \to 1} x^4 = \lim_{x \to 1} \frac1x = 1 $

for the second one if you choose $\delta = \frac{\epsilon}{2}$ then $x > \frac12$ since $x$ is near $1$ and you have $$|x-1| < \frac{\epsilon}{2} \implies |x-1| < x \epsilon \implies|\frac1x -1|< \epsilon$$

for the first one note that $x^4-1 = (x-1)(x+1)(x^2+1)$ and that $\frac{1}{(x+1)(x^2+1)} > \frac{1}{15}$

you could then try $\delta = \frac{1}{15}\epsilon$

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Hint:$$\left|\frac{x^5+1}{x}-2\right|=\left|\frac{x^5-2x+1}{x}\right|=\left|\frac{(x-1) \left(x^4+x^3+x^2+x-1\right)}{x}\right|=\delta\left|\frac{x^4+x^3+x^2+x-1}{x}\right|$$

Now find an upper bound for $\left|\frac{x^4+x^3+x^2+x-1}{x}\right|$

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