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Let $M>0$ a constant and $f:\mathbb{R}\to\mathbb{R}$ a function such that $|f(x)|\geq M$ for all $x\in\mathbb{R}$. Calculate the next limit: $$\lim_{x\to1} \dfrac{x-1}{f(x)}$$

I'm really stuck in this exercise don't know how to set the inequality to apply the squeeze theorem.

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  • $\begingroup$ should it not just be zero? $\endgroup$ – abel Oct 2 '15 at 19:13
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    $\begingroup$ $|f(x)|\ge M$, so $\dfrac 1{|f(x)|} \le \dfrac 1 M$, and finally $\left|\dfrac{x-1}{f(x)}\right| \le \dfrac{|x-1|} M \to 0$ as $x\to1$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 2 '15 at 19:17
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Hint

Since $|f(x)|\geq M$,

$$0\leq \left|\frac{x-1}{f(x)}\right|\leq \left|\frac{x-1}{M}\right|$$

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since $0 \leqslant |\frac{x-1}{f(x)}| \leqslant \frac{|x-1|}{M}$ and $\frac{|x-1|}{M} \to 0$

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