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Can someone verify I've done this right? Lim x->0 of (sin^2 (5x))/7x^2. I've come up with 25/49.
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\[\lim_{x \rightarrow 0}\frac{\sin(5x) \cdot \sin(5x)}{7 \cdot x \cdot x}=\frac{25}{7}\lim_{x \rightarrow 0}\frac{\sin(5x)}{5x} \cdot \frac{\sin(5x)}{5x} \]
the 49 is a mistake
yes it is just 7 on bottom
what the talented myininaya said
Devilry! I accidently multiplied 7x twice. I shouldn't be trusted with algebra. Thanks.
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no one should be trusted with algebra
its all to get us all
i'm jk mistakes happen lol
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