Question

find the answer for my exam preparation
lim x→0 (cosecx - cotx)
​

Answer 1

Given :-

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$\tt\longrightarrow{\lim_{x \rightarrow 0}\: (cosec\: x - cot\: x)}$

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Solution :-

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As we know that

$\: \: \: \: \: \: \: \: \bullet\bf\: \: \: {cosec\: x = \dfrac{1}{sin\: x}}$

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$\: \: \: \: \: \: \: \: \bullet\bf\: \: \: {cot\: x = \dfrac{cos\: x}{sin\: x}}$

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So we can write

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$\bf\implies{\lim_{x \rightarrow 0}\: \bigg( \dfrac{1}{sin\: x} - \dfrac{cos\: x}{sin\: x} \bigg) }$

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$\tt:\implies{\lim_{x \rightarrow 0}\: \bigg( \dfrac{1 - cos\: x}{sin\: x} \bigg)}$

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Now, dividing both numerator and denominator by x.

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$\tt:\implies{\lim_{x \rightarrow 0}\: \bigg( \dfrac{\: \: \dfrac{1 - cos\: x}{x}}{\dfrac{sin\: x}{x}} \bigg) }$

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We know that

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$\: \: \: \: \: \: \: \: \bullet\bf\: \: \: {\lim_{x \rightarrow 0}\: \dfrac{1 - cos\: x}{x} = 0}$

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$\: \: \: \: \: \: \: \: \bullet\bf\: \: \: {\lim_{x \rightarrow 0} \: \dfrac{sin\: x}{x} = 1}$

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$\tt:\implies{\dfrac{0}{1}}$

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$\tt:\implies{0}$

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Hence,

• Limit of the given function is 0.

Answer 2

Step-by-step explanation:

$\large\bf{\underline{\underline{We\ have:-}}}$

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$\bf : \implies {\displaystyle \lim_{\bf x \to 0}\ \bf (cosec\ x\ -\ cot\ x)}$

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$\large\bf{\underline{\underline{Solution:-}}}$

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❏ We know that:-

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$\bf : \implies {cosec\ x\ =\ \dfrac{1}{sin\ x}} \\ \\ \\ \bf : \implies {cot\ x\ =\ \dfrac{cos\ x}{sin\ x}}$

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❏ We can write it as:-

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$\bf : \implies {\displaystyle \lim_{\bf x \to 0}\ \bf \bigg (\dfrac{1}{sin\ x}\ -\ \dfrac{cos\ x}{sin\ x} \bigg)} \\ \\ \\ \bf : \implies {\displaystyle \lim_{\bf x \to 0}\ \bf \dfrac{1\ -\ cos\ x}{sin\ x}}$

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❏ By putting x = 0:-

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$\bf : \implies {\dfrac{1\ -\ cos\ 0}{sin\ 0}} \\ \\ \\ \bf : \implies {\dfrac{1\ -\ 1}{0}} \\ \\ \\ \bf : \implies {\dfrac{0}{0}}$

● Since, it is coming as 0/0.

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❏ Now we will simplify it:-

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$\bf : \implies {\displaystyle \lim_{\bf x \to 0}\ \bf \dfrac{1\ -\ cos\ x}{sin\ x}} \\ \\ \\ \bf : \implies {\displaystyle \lim_{\bf x \to 0}\ \bf \dfrac{1\ -\ cos\ x}{sin\ x}\ \times\ \dfrac{1\ +\ cos\ x}{1\ +\ cos\ x}} \\ \\ \\ \bf : \implies {\displaystyle \lim_{\bf x \to 0}\ \bf \dfrac{1^2\ -\ cos^2\ x}{sin\ x\ (1\ +\ cos\ x)}} \\ \\ \\ \bf : \implies {\displaystyle \lim_{\bf x \to 0}\ \bf \dfrac{1\ -\ cos^2\ x}{sin\ x\ (1\ +\ cos\ x)}} \\ \\ \\ \bf : \implies {\displaystyle \lim_{\bf x \to 0}\ \bf \dfrac{sin^2\ x}{sin\ x\ (1\ +\ cos\ x)}} \\ \\ \\ \bf : \implies {\displaystyle \lim_{\bf x \to 0}\ \bf \dfrac{sin\ x}{(1\ +\ cos\ x)}}$

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❏ By putting x = 0:-

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$\bf : \implies {\dfrac{sin\ 0}{(1\ +\ cos\ 0)}} \\ \\ \\ \bf : \implies {\dfrac{0}{1\ +\ 1}} \\ \\ \\ \bf : \implies {\dfrac{0}{2}} \\ \\ \\ \bf : \implies {\underline{\fbox{\pink{\bf 0.}}}}\bigstar$

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$\sf \therefore {\underline{Hence,\ the\ limit\ of\ the\ given\ function\ is\ \bf 0.}}$