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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Can someone verify I've done this right? Lim x->0 of (sin^2 (5x))/7x^2. I've come up with 25/49.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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

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what the talented myininaya said
Devilry! I accidently multiplied 7x twice. I shouldn't be trusted with algebra. Thanks.
no one should be trusted with algebra
its all to get us all
i'm jk mistakes happen lol

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