Find the value of the limit $\lim _ { x \rightarrow 0 } \frac { \cos x - 1 } { 2 x }$ .
A)3
B)0
C) $\frac { 1 } { 3 }$
D)4
E) $\frac { 1 } { 4 }$
F)2
G) $\frac { 1 } { 2 }$
H)1

Question

Find the value of the limit $\lim _ { x \rightarrow 0 } \frac { \cos x - 1 } { 2 x }$ .
A)3
B)0
C) $\frac { 1 } { 3 }$ 
D)4
E) $\frac { 1 } { 4 }$ 
F)2
G) $\frac { 1 } { 2 }$ 
H)1

Quizplus · Accepted Answer

The limit $\lim _ { x \rightarrow 0 } \frac { \cos x - 1 } { 2 x }$ can be solved by applying L'Hôpital's Rule, which allows us to differentiate the numerator and the denominator separately when the limit initially gives an indeterminate form like 0/0. Differentiating the numerator gives $-\sin x$ and differentiating the denominator gives 2. Thus, the limit becomes $\lim _ { x \rightarrow 0 } \frac { -\sin x } { 2 }$ = $\frac { -\sin(0) } { 2 }$ = 0.

Find the Value of the Limit lim⁡x→0cos⁡x−12x\lim _ { x \rightarrow 0 } \frac { \cos x - 1 } { 2 x }limx→0​2xcosx−1​

Find the Value of the Limit $\lim _ { x \rightarrow 0 } \frac { \cos x - 1 } { 2 x }$