What is the limit of $\frac{sin(sin\ x)}{x}$ as $x$ approaches infinity?

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The limit of the expression \(\frac{\sin(\sin x)}{x}\) as \(x\) approaches infinity is definitively 0. As \(x\) increases, the denominator grows without bound while the numerator, \(\sin(\sin x)\), remains bounded between -1 and 1. Therefore, the overall limit converges to 0 due to the dominance of the denominator in the fraction.

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Homework Statement


\mathop{\lim}\limits_{a \to \infty} \frac{sin(sin\ x )}{x}

The Attempt at a Solution



I was thinking about starting off by multiplying by sin(x) over sin(x)

\mathop{\lim}\limits_{a \to \infty} \frac{sin(sin\ x )}{x} * \frac{sin\ x }{sin \ x}Don't know if that would help...

Thanks
 
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planauts said:

Homework Statement


\mathop{\lim}\limits_{a \to \infty} \frac{sin(sin\ x )}{x}

The Attempt at a Solution



I was thinking about starting off by multiplying by sin(x) over sin(x)

\mathop{\lim}\limits_{a \to \infty} \frac{sin(sin\ x )}{x} * \frac{sin\ x }{sin \ x}


Don't know if that would help...

Thanks

What is a? Do you mean x? And (sinx)/x goes to 1 only if x is approaching 0. So I'm not sure this would work.
 
-1\leq\sin(x)\leq 1, so the value inside the parentheses restricts the outside sin() function, while the denominator grows without bound
 

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