Question

✬ Hey there!

Evaluate the given limit's :

$\tt \bull \: \tt \tt \: lim_{x \rightarrow 0} \bigg( \csc(x) - \cot(x) \bigg)$
$\bull \: \tt \: lim_{x \rightarrow 0} \bigg( \dfrac{ \cos(x) }{ \pi - x} \bigg)$


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Answer 1

Answer:

1)0

2)1/π

Step-by-step explanation:

Using formula:-

• Cosec X=1/Sin X
• Cot X=Cos X/Sin X
• Sin² X+Cos² X=1
• Sin0°=0
• Cos0°=1

Answer 2

GIVEN :-

$\sf\: \sf \sf\: lim_{x \rightarrow 0} \bigg( \cosec(x) - \cot(x) \bigg)$

$\sf\: lim_{x \rightarrow 0} \bigg( \dfrac{ \cos(x) }{ \pi - x} \bigg)$

SOLUTION :-

$\sf \bull 1 \: \: solution$

$\sf\: \sf \sf\: lim_{x \rightarrow 0} \bigg( \cosec(x) - \cot(x) \bigg)$

$= \sf\: \sf \sf\: lim_{x \rightarrow 0} \bigg( \frac{1}{ \sin(x) } - \frac{ \cos(x) }{ \sin(x) } \bigg)</p><p>$

$= \sf\: \sf \sf\: lim_{x \rightarrow 0} \bigg( \frac{1 - \cos(x) }{ \sin(x) } \bigg)</p><p>$

$\sf \: Formula -$

$\green{ \sf \bull \:( 1 - \cos(x) ) = 2 \sin^{2} \bigg( \frac{x}{2} \bigg)}$

$\green{ \sf \bull \: sinx = 2 \sin \bigg( \frac{x}{2} \bigg) .\cos \bigg( \frac{x}{2} \bigg) }$

$= \sf\: \sf \sf\: lim_{x \rightarrow 0} \bigg( \frac{2 \sin^{2} ( \frac{x}{2} ) }{ 2\sin( \frac{x}{2} ) . \cos( \frac{x}{2} ) } \bigg)</p><p>$

$= \sf\: \sf \sf\: lim_{x \rightarrow 0} \bigg( \frac{\sin ( \frac{x}{2} ) }{ \cos( \frac{x}{2} ) } \bigg)$

$= \sf\: \sf \sf\: lim_{x \rightarrow 0} \: \: \tan( \frac{x}{2} )$

$\sf \: Now \: putting \: th e\: limits$

$\sf \: = \tan(0)$

$= 0$

$\orange{\sf\: \sf \sf\: lim_{x \rightarrow 0} \bigg( \cosec(x) - \cot(x) \bigg) = 0</p><p> }$

$\sf \bull 2 \: \: solution$

$\sf\: lim_{x \rightarrow 0} \bigg( \dfrac{ \cos(x) }{ \pi - x} \bigg)$

$\sf \: putting \: the \: limits$

$= \sf\: \bigg( \dfrac{ \cos(0) }{ \pi - 0} \bigg)$

$= \frac{1}{\pi}$

.................

$\orange{\sf\: lim_{x \rightarrow 0} \bigg( \dfrac{ \cos(x) }{ \pi - x} \bigg) = \frac{1}{\pi} }$

$\sf \red {\: I \: think \: the \: limits \: of \: 2nd \: question \: is \: \pi}$

So

$\sf\: lim_{x \rightarrow \pi} \bigg( \dfrac{ \cos(x) }{ \pi - x} \bigg)$

Here denominator become zero as we putting the value of x

Then. appling L hospital rule

which is differencate the denominator

$= \sf\: lim_{x \rightarrow \pi} \bigg( \dfrac{ \frac{d}{dx}( \cos(x)) }{ \frac{d}{dx}( \pi - x)} \bigg)$

$= \sf\: lim_{x \rightarrow \pi} \bigg( \dfrac{ - \sin(x) }{ 0 - 1} \bigg)$

$= \sf\: lim_{x \rightarrow \pi} \: \sin(x)$

Putting the limits

$= \sin(\pi)$

$= 0$

$\orange{\sf\: lim_{x \rightarrow \pi} \bigg( \dfrac{ \cos(x) }{ \pi - x} \bigg) = 0}$