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I was just wondering about this query:

lim((cosx/x^2)-(sinx)/x^3))

x->0

My teacher is telling me to use L'hopital, but the problem is that

the first part (cosx)/(x^2) isn't a "0/0", cosx-> 1 for x->0

So what should I do, i know the right answer is -1/3, but i need to prove it.

Plz. Help Me

Thanks

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# Lim((cosx/x^2)-(sinx)/x^3)) x->0 L'hopital

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