Question

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

please give it correct and fastt...please..​

Answer 1

Answer:

Step-by-step explanation:

At x = 0,

the value of the given function takes the form.

see the attachment

Answer 2

Step-by-step solution:-

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

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