Question

$\huge \underline{\sf\red{question}}$
$\bigstar \: \underline\bold{evaluate \: the \: following \: limit - }$
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Answer 1

Topic :-

Limits

To Find :-

$\displaystyle \sf{\lim_{x\to a}\dfrac{\cos x- \cos a}{\cot x- \cot a}}$

Concept Used :-

L' Hôpital's Rule

It is applicable while calculating limits of indeterminate forms of type 0/0 or ∞/∞.

$\displaystyle \sf {If\:\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0\:or\:\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=\infty,}$

$\displaystyle \sf{then\:\lim_{x\to a}\dfrac{f(x)}{g(x)}=\lim_{x\to a}\dfrac{f'(x)}{g'(x)}}$

$\displaystyle \sf{provided\:that\:g'(x)\neq 0\:and\:limit \:\lim_{x\to a}\dfrac{f'(x)}{g'(x)}\:exists.}$

Solution :-

$\displaystyle \sf{\lim_{x\to a}\dfrac{\cos x- \cos a}{\cot x- \cot a}}$

Put x = a to check its indeterminate form.

$\sf{\dfrac{\cos a- \cos a}{\cot a- \cot a}}$

$\sf{\dfrac{0}{0}}$

Indeterminate form of given expression is 0/0, hence, L' Hôpital's Rule can be applied here.

Applying L' Hôpital's Rule,

$\displaystyle \sf{\lim_{x\to a}\dfrac{f(x)}{g(x)}=\lim_{x\to a}\dfrac{f'(x)}{g'(x)}}$

$\displaystyle \sf{\lim_{x\to a}\dfrac{\cos x- \cos a}{\cot x- \cot a}=\lim_{x\to a}\dfrac{\dfrac{d}{dx}(\cos x- \cos a)}{\dfrac{d}{dx}(\cot x- \cot a)}}$

$\displaystyle \sf{\lim_{x\to a}\dfrac{\dfrac{d}{dx}(\cos x- \cos a)}{\dfrac{d}{dx}(\cot x- \cot a)}=\lim_{x\to a}\dfrac{\dfrac{d}{dx}(\cos x)- \dfrac{d}{dx}(\cos a)}{\dfrac{d}{dx}(\cot x)- \dfrac{d}{dx}(\cot a)}}$

$\sf {\left(\because \dfrac{d(f\pm g)}{dx}=\dfrac{d(f)}{dx}\pm \dfrac{d(g)}{dx} \right)}$

'cos a' and 'cot a' act as constant with respect to 'x'.

$\displaystyle \sf{\lim_{x\to a}\dfrac{\dfrac{d}{dx}(\cos x)-0}{\dfrac{d}{dx}(\cot x)-0}}$

$\sf {\left(\because \dfrac{d(k)}{dx}=0,where\:k\:is\:a\:constant.\right)}$

$\displaystyle \sf{\lim_{x\to a}\dfrac{-\sin x}{\dfrac{d}{dx}(\cot x)}}$

$\sf {\left(\because \dfrac{d(\cos x)}{dx}=-\sin x \right)}$

$\displaystyle \sf{\lim_{x\to a}\dfrac{-\sin x}{-\csc^2x}}$

$\sf {\left(\because \dfrac{d(\cot x)}{dx}=-\csc^2x \right)}$

$\displaystyle \sf{\lim_{x\to a}\dfrac{\sin x}{\dfrac{1}{\sin^2x}}}$

$\sf{\left(\because \csc x=\dfrac{1}{\sin x}\right)}$

$\displaystyle \sf{\lim_{x\to a}\sin^3x}$

Put value of the limit that is x = a.

$\sf{\sin^3a}$

Answer :-

$\displaystyle \underline{\boxed{\sf{\lim_{x\to a}\dfrac{\cos x- \cos a}{\cot x- \cot a}=sin^3a}}}$

Answer 2

❏Question:-

$\begin{gathered} \sf \: find \: the \: value \: of \: \: \red{\lim_{x \to a} } \bigg \{\green{ \frac{ \cos x - \cos a}{ \cot x - \cot a} } \bigg \}\\ \end{gathered}$

❏Answer :-

$\boxed{ \red{\lim_{x \to a} } \bigg \{\green{ \frac{ \blue{\cos x - \cos a}}{ \cot x - \cot a} } \bigg \} \: \: = { \sin}^{3} a} \: \:$

❏Solution :-

We have ,

$\begin{gathered} \implies \pink{ \lim_{x \to a}} \bigg \{ \blue{ \frac{ \cos x - \cos a}{ \cot x - \cot a} }\bigg \} \\ \\ \bf \: Taking \: limit \: \\ \\ \implies \sf \frac{ \cos a - \cos a}{ \cot a - \cot a} \\ \\ \implies \: \frac{0}{0} \: \{ \sf \green{ indeterminate \: form }\} \\ \\ \bf \: hence \: \\ \bf \red{Apply \: L } \: - Hospital \: rule \: \\ \to \sf \: differentiate \: numenator \: and \: denomenator \: \\ \sf \: \: \: \: \: \: both \: \\ \\ \to \: \red{\lim_{x \to a} }\green{ \frac{ \cos x - \cos a}{ \cot x - \cot a} } \\ \: \\ \bf after \: differentiation \: \\ \\ \to \: \blue{\lim_{x \to a} }\pink{ \frac{ - \sin x }{ - { \csc}^{2}x } } \\ \because \: \boxed{ \csc \theta = \frac{1}{ \sin \theta} } \\ \therefore \\ \to \: \purple{\lim_{x \to a} } \: \: \red{ \frac{ \sin x }{ \frac{1}{ { \sin}^{2}x } } }\: \\ \\ \to \: {\lim_{x \to a} } \: \: \red{ { \sin}^{3} x} \: \\ \bf \red{ Taking \: limit \: } \\ \\ \to \: { \sin}^{3} a \\ \\ \boxed{ \red{\lim_{x \to a} } \bigg \{\green{ \frac{ \cos x - \cos a}{ \cot x - \cot a} } \bigg \} \: \: = { \sin}^{3} a}\end{gathered}$

$\sf{\blue{\underline{\underline{\orange{HENCE\; VERIFIED}}}}}$

• $\sf @AbhinavRocks10$