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(more precisely, the largest $A > 0$ and the smallest $b>0$) and Suppose $0 < \varepsilon < 1$.

Statement: The function $\frac{\sin x}{x}$ tends to the limit $1$ as $x \to 0^+$ because, given $\varepsilon > 0$, there exists $d > 0$ depending on $\varepsilon$ such that

$$\left|\frac{\sin x}{x} - 1\right| < \varepsilon$$

whenever $0 < x < d$.


This is also a practice question for my exam coming up.

Using the $\varepsilon$-$\delta$ definition of a limit I have simply replaced $d$ by $A\varepsilon^b$ in the inequality:

$$0<|x-0|< A\varepsilon^b$$

$$|x|< A\varepsilon^b$$

$$\ln |x| = b \ln(A\varepsilon)$$

And then solving for $A$ and $b$ from there. However, I am pretty sure this is wrong, but I am really struggling to find another way to solve this?

Any direction would be greatly appreciated!


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You have to find the largest A and the smallest b such that WHICH STATEMENT is correct? – g.castro Jun 4 '12 at 12:34
Sorry, I have edited the post. – JackReacher Jun 5 '12 at 4:19

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