Question

lim x tends to 0( x^3.cotx)/1-cos x

Answer 1

Rationalise it than your denominator becomes sin square x then multiply and divide by x You will get answer

Answer 2

$\lim_{x \to 0} \dfrac{x^3.\ cotx}{1-\cos x}=2$

Step-by-step explanation:

We have,

$\lim_{x \to 0} \dfrac{x^3.\ cotx}{1-\cos x}$

To find, the value of $\lim_{x \to 0} \dfrac{x^3.\ cotx}{1-\cos x}=?$

∴ $\lim_{x \to 0} \dfrac{x^3.\ cotx}{1-\cos x}$

Rationalising numerator and deniminator, we get

$\lim_{x \to 0} \dfrac{x^3.\ cotx}{1-\cos x}\times\dfrac{1+\cos x}{1+\cos x}$

$=\lim_{x \to 0} \dfrac{x^3.\ cotx(1+\cos x)}{1^2-\cos^2 x}$

$=\lim_{x \to 0} \dfrac{x^3.\ cosx(1+\cos x)}{\sin x\sin^2 x}$

$=\lim_{x \to 0} \dfrac{x^3.\ cosx(1+\cos x)}{\sin^3 x}$

$=\lim_{x \to 0} \dfrac{\ cosx(1+\cos x)}{\dfrac{\sin^3 x}{x^3} }$

$=\lim_{x \to 0} \dfrac{\ cosx(1+\cos x)}{(\dfrac{\sin x}{x})^3}$

$=\lim_{x \to 0} \dfrac{\ cosx(1+\cos x)}{1}$

[ ∵ $=\lim_{x \to 0} \dfrac{\sin x}{x}=1$]

Put x 0, we get

$=\dfrac{\ cos0(1+\cos 0)}{1}$

[ ∵$\cos0=1$]

$=\dfrac{1(1+1)}{1}=2$

Hence, $\lim_{x \to 0} \dfrac{x^3.\ cotx}{1-\cos x}=2$