displaystyle limx → 0 ( √1 - cos 2x/√2 x) is

Question

displaystyle limx → 0 ( √1 - cos 2x/√2 x) is (A) 1 (B) -1 (C) 0 (D) does not exist

Tardigrade Admin · Accepted Answer

lim(√1- cos 2x/√2 x) ⇒ lim (√1-(1-2 sin2 x)/√2 x) displaystyle limx →0 (√2 sin2 x/√2 x) ⇒ limx →0 (| sin x|/x) The limit of above does not exist as LHS = -1 ≠ RHL = 1

Q. $x \to 0 lim \frac{1 - cos 2 x}{2 x}$ is

1

-1

0

does not exist

$lim \frac{1 - c o s 2 x}{2 x} \Rightarrow lim \frac{1 - ( 1 - 2 s i n ^{2} x )}{2 x}$
$x \to 0 lim \frac{2 sin ^{2} x}{2 x} \Rightarrow x \to 0 lim \frac{∣ sin x ∣}{x}$
The limit of above does not exist as
LHS = -1 $\neq =$ RHL = 1

Q. x→0lim​2​x1−cos2x​​ is

1

-1

0

does not exist

lim2​x1−cos2x​​⇒lim2​x1−(1−2sin2x)​​ x→0lim​2​x2sin2x​​⇒x→0lim​x∣sinx∣​ The limit of above does not exist as LHS = -1 = RHL = 1

Q. $x \to 0 lim \frac{1 - cos 2 x}{2 x}$ is

$lim \frac{1 - c o s 2 x}{2 x} \Rightarrow lim \frac{1 - ( 1 - 2 s i n ^{2} x )}{2 x}$
$x \to 0 lim \frac{2 sin ^{2} x}{2 x} \Rightarrow x \to 0 lim \frac{∣ sin x ∣}{x}$
The limit of above does not exist as
LHS = -1 $\neq =$ RHL = 1