Question

Help Me To Find the Value of the Limit ... Please
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Answer 1

Answer:

-sin (2y) / (2y), if y ≠ 0

-1, if y = 0

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Step-by-step explanation:

As x approaches y, the expression tends to the form 0/0.  L'Hopital's Rule is useful here.  It tells use that the limit will be the same if we differentiate the numerator and denominator separately with respect to x. So...

$\displaystyle\lim_{x\rightarrow y}\frac{\cos^2x-\cos^2y}{x^2-y^2}\\ \\=\lim_{x\rightarrow y}\frac{-2\cos x \sin x}{2x}\\ \\=\lim_{x\rightarrow y}\frac{-\sin 2x}{2x}\\ \\=\begin{cases}-\frac{\sin 2y}{2y} &\text{if\ }y\neq 0\\1&\text{if\ }y=0.\end{cases}$