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For what values a and b is the limit as x approaches 0 of (a-cos(bx))/x^2 = 50
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\[\lim_{x\to0}\frac{a-\cos bx}{x^2}=\frac{a-1}{0}\] Clearly, this is undefined for \(a\not=1\). But since the limit must exist, we require that the numerator also approaches 0 so that we can apply L'Hopital's rule. This means we must have \(a=1\). Applying L'Hopital's rule yields \[\lim_{x\to0}\frac{b\sin bx}{2x}=\frac{0}{0}\] Apply the rule again: \[\lim_{x\to0}\frac{b^2\cos bx}{2}=\frac{b^2}{2}\] The limit is known to be 50, so you must have \[\frac{b^2}{2}=50\]
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