$\lim_{x \rightarrow 0} \frac{\sin^{2}\left(\pi \cos^{4} x\right)}{x^{4}}$ is equal to:

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Limits

lim_{x \to 0} \frac{s i n ^{2} ( π c o s ^{4} x )}{x ^{4}}

is equal to:

Answer: (B)

For small

x

cos x = 1 - \frac{x ^{2}}{2} + \dots

, so

cos^{4} x = 1 - 2 x^{2} + \dots

and

π cos^{4} x = π - 2 π x^{2} + \dots

. Then

sin (π cos^{4} x) = sin (π - 2 π x^{2}) = sin (2 π x^{2}) \approx 2 π x^{2}

. Hence

\frac{sin ^{2} ( π cos ^{4} x )}{x ^{4}} \to \frac{( 2 π x ^{2} ) ^{2}}{x ^{4}} = 4 π^{2}

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · 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 →