$\lim _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$ is equal to :

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Limits

lim_{x \to 0} \frac{c o s ( s i n x ) - c o s x}{x ^{4}}

is equal to :

Answer: (C)

Using the identity

cos A - cos B = - 2 sin (\frac{A + B}{2}) sin (\frac{A - B}{2})

lim_{x \to 0} \frac{2 s i n ( \frac{x + s i n x}{2} ) s i n ( \frac{x - s i n x}{2} )}{x ^{4}}

Using

sin θ \approx θ

for small

θ

, this becomes:

lim_{x \to 0} \frac{( x + s i n x ) ( x - s i n x )}{2 x ^{4}} = lim_{x \to 0} \frac{x ^{2} - s i n ^{2} x}{2 x ^{4}}

Applying L'Hôpital's rule twice:

lim_{x \to 0} \frac{2 x - s i n 2 x}{8 x ^{3}} = lim_{x \to 0} \frac{2 - 2 c o s 2 x}{24 x ^{2}} = lim_{x \to 0} \frac{4 ( 1 - c o s 2 x )}{24 ( 4 x ^{2} )} \cdot 4 = \frac{4 \cdot \frac{1}{2}}{24} = \frac{1}{6}

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Single correct option

Let $f : R \to (0, \infty)$ be a twice differentiable function such that $f (3) = 18, f^{'} (3) = 0$ and $f^{''} (3) = 4$ .
Then …
JEE Main 2026View solution →
If $lim_{x \to 0} \frac{e ^{(a - 1) x} + 2 c o s b x + ( c - 2 ) e ^{- x}}{x c o s x - l o g _{e} ( 1 + x )} = 2$ , then $a^{2} + b^{2} + c^{2}$ is equal to :
JEE Main 2026View solution →
The value of
$lim_{x \to 0} \frac{l o g _{e} ( s e c ( e x ) \cdot s e c ( e ^{2} x ) \cdot \dots \cdot s e c ( e ^{10} x ) )}{e ^{2} - e ^{2 c o s x}}$
is equal to
JEE Main 2026View solution →
If $lim_{x \to 2} \frac{s i n ( x ^{3} - 5 x ^{2} + a x + b )}{( x - 1 - 1 ) l o g _{e} ( x - 1 )} = m$ , then $a + b + m$ is equal to:
JEE Main 2026View solution →
The product of all possible values of $α$ , for which $lim_{x \to 0} (\frac{1 - c o s ( α x ) c o s (( α + 1 ) x ) c o s (( α + 2 ) x )}{s i n ^{2} (( α + 1 ) x )}) = 2$ , is :
JEE Main 2026View solution →
Let $f (x) = lim_{y \to 0} \frac{( 1 - c o s ( x y )) t a n ( x y )}{y ^{3}}$ . Then the number of solutions of the equation $f (x) = sin x$ , $x \in R$ is:
JEE Main 2026View solution →