Question

Lim x tends to a cosx-cosa/cotx-cota
Plzz solve it step wise.. .. .. .. .. . Fast

Answer 1

method 1 :- $\textbf{ it can be easily solved with help of \underline{L- Hospital rule}}$

Given, $\displaystyle{lim_{x\to\ a}\frac{cosx-cosa}{cotx-cota}}$
Let's check form of limit , put x = a
we get , 0/0 . so we can use L - Hospital rule because L - Hospital can be applied for 0/0 and ∞/∞ .

now, differentiate numerator and denominator separately.
then, $=lim_{x\to a}\frac{-sinx}{-cosec^2x}$
$=lim_{x\to\ a }sin^3x$
now put x = a
then,answer is $\bf{sin^3a}$

method 2 :-
$Let,f(x)=\frac{cosx-cosa}{cotx-cota}\\\\=\frac{cosx-cosa}{\frac{cosx}{sinx}-\frac{cosa}{sina}}\\\\=\frac{(cosx-cosa)sinx.sina}{sin(a-x)}\\\\=\frac{2sin\frac{x+a}{2}sin\frac{a-x}{2}sinx.sina}{2sin\frac{a-x}{2}cos\frac{a-x}{2}}\\\\=\frac{sin\frac{a+x}{2}sinx.sina}{cos\frac{a-x}{2}}$

now, put it in limit ,
so, $Lim_{x\to\ a} \frac{cosx-cosa}{cotx-cota}=lim_{x\to\ a} \frac{sin\frac{a+x}{2}sinx.sina}{cos\frac{a-x}{2}}$
now put x = a
then, answer is $\bf{sin^3a}$

Answer 2

Answer:

sin^3 (a)

Step-by-step explanation:

Lim x>a (cos x cos a / cot x cot a)

= Lim x>a (cos x cos a) / [(cos x sin a cosa sin x)/(sin x sin a)]

= Lim x>a (2sin((x-a)/2)sin((x+a)/2)) / [(sin(ax))/(sin x sin a)]  

= sin^3 a.