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Evaluate: $$\lim_\limits{x\to \pi/4} \frac {2-\csc^2 x}{1-\cot x}$$

My Attempt: \begin{align}\lim_\limits{x\to \pi/4} \frac {2-\csc^2 x}{1-\cot x}&=\lim_{x\to \pi/4} \frac {2-\csc^2 x}{1-\cot x}\\\\ &=\lim_{x\to \pi/4} \frac {2-\frac {1}{\sin^2 x}}{1-\frac {\cos x}{\sin x}}\\\\ &=\lim_{x\to \pi/4} \frac {2\sin^2 x - 1}{\sin^2 x - \sin x\cdot\cos x}\end{align}

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  • $\begingroup$ its far more readable $\lim\limits_{x\to\pi/4}...$ than what you had written. $\endgroup$ – Masacroso Jul 18 '17 at 10:05
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    $\begingroup$ If you simplify the fraction it becomes $\cot x+1\to 2$ as $x\to \frac{\pi}{4}$ $\endgroup$ – Raffaele Jul 18 '17 at 10:06
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We use the trigonometric identity $1 + \cot^2x = \csc^2x$ to simplify the expression. \begin{align*} \lim_{x \to \frac{\pi}{4}} \frac{2 - \csc^2x}{1 - \cot x} & = \lim_{x \to \frac{\pi}{4}} \frac{2 - (1 + \cot^2x)}{1 - \cot x}\\ & = \lim_{x \to \frac{\pi}{4}} \frac{1 - \cot^2x}{1 - \cot x}\\ & = \lim_{x \to \frac{\pi}{4}} \frac{(1 + \cot x)(1 - \cot x)}{1 - \cot x}\\ & = \lim_{x \to \frac{\pi}{4}} (1 + \cot x)\\ & = 1 + 1\\ & = 2 \end{align*}

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Note first that $\sin^2 x+\cos^2=1$ hence we obtain $$2-\frac1{\sin^2x}=\frac{2\sin^2x-1}{\sin^2x}=\frac{\sin^2x-\cos^2x}{\sin^2x}=1-\cot^2x=(1-\cot x)(1+\cot x)$$

hence $$\lim_{x\to\frac{\pi}4}\frac{2-\frac1{\sin^2x}}{1-\cot x}=\lim_{x\to\frac{\pi}4}\left(1+\cot x\right)=2.$$

You could also use a L'Hopital's rule.

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Considering $$ A=\dfrac {2-\csc^2( x)}{1-\cot(x)}$$ let $x=y+\frac \pi 4$ to get (after simplification) $$A=\frac{2 \cos (y)}{\sin (y)+\cos (y)}$$ and now, look for the limit when $y\to 0$.

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