Question

Evaluate this limit , lim x tends to 0 cot theta - cosec theta .​

Answer 1

GiveN :-

• A limit is given to us and we need to evaluate that limit.
• $\displaystyle\sf \lim_{\theta \to 0 } cot\theta - cosec \theta$

To Do :-

• To evaluate the limit .

SolutioN :-

Basically if we directly substite $\theta$ = 0 in the limit . It will lead to indeterminate form . Because the value of $\sf cot\ 0^o = \infty$ and the value of $\sf cosec\ 0^o= \infty$

We will simply it . Let's do it !

$\implies \displaystyle\sf\lim_{\theta \to 0} cot\theta - cosec\theta$

$\implies \displaystyle\sf\lim_{\theta \to 0} \dfrac{cos\theta}{sin\theta}-\dfrac{1}{sin\theta}$

$\implies \displaystyle\sf\lim_{\theta \to 0} \dfrac{cos\theta-1}{sin\theta}$

$\implies \displaystyle\sf\lim_{\theta \to 0}\dfrac{(cos\theta-1)(cos \theta +1)}{sin\theta(cos\theta+1)}$

$\implies \displaystyle\sf\lim_{\theta \to 0}\dfrac{cos^2\theta-1}{sin\theta(cos\theta+1)}$

$\implies \displaystyle\sf\lim_{\theta \to 0} \dfrac{-1(1-cos^2\theta)}{sin\theta(cos\theta+1)}$

$\implies \displaystyle\sf\lim_{\theta \to 0}\dfrac{\cancel{-sin^2\theta}}{ \cancel{sin\theta}(cos\theta+1)}$

$\implies \displaystyle\sf\lim_{\theta \to 0}\dfrac{-sin\theta}{1+cos\theta}$

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$\red{\bigstar}\underline{\textsf{ Now put the value of limit. }}$

$\sf\to \dfrac{-sin 0^{\circ}}{1+cos0^{\circ}}\\\\\sf\to \dfrac{-0}{1+1}\\\\\sf\to \dfrac{0}{2} \\\\\sf\to \pink{\textsf{\textbf{ 0 }}} \qquad\qquad\bigg\lgroup \red{\tt Required \ Answer }\bigg\rgroup$

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