Math Help - Limits Problem, can someone help with this?

1. Originally Posted by Also sprach Zarathustra
Here something useful...

$lim_{x\to 0} \frac{sin x}{x}=1$

$lim_{x\to 0} \frac{sin 4x}{9x}=\frac{4}{9}lim_{x\to 0} \frac{sin 4x}{4x}$

Try apply this also for b.
How did you go from sin4x/9x to 4/9(sin4x/4x?) I don't get how you get the 4x on the bottom

2. Originally Posted by Nikhiln25
How did you go from sin4x/9x to 4/9(sin4x/4x?) I don't get how you get the 4x on the bottom

$\frac{sin4x}{9x}=\frac{sin4x}{4x\cdot \frac{9}{4}}=\frac{4}{9} \frac{sin4x}{4x}$

yes?

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