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

Question

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

Quizplus · Accepted Answer

The limit $\lim _ { x \rightarrow 0 ^ { + } } \frac { \cos x - 1 } { x ^ { 2 } }$ can be evaluated using L'Hôpital's Rule, which states that if the limit of functions $f(x)$ and $g(x)$ as $x$ approaches a point $c$ results in an indeterminate form like $0/0$ or $\infty/\infty$, then the limit of their derivatives can be taken instead. Applying this rule, we differentiate the numerator and the denominator separately:Numerator's derivative: $\frac{d}{dx}(\cos x - 1) = -\sin x$Denominator's derivative: $\frac{d}{dx}(x^2) = 2x$Thus, the limit becomes $\lim _ { x \rightarrow 0 ^ { + } } \frac { -\sin x } { 2x }$.Applying L'Hôpital's Rule again or recognizing this as the derivative of $\sin x$ at $x = 0$, we get:$\lim _ { x \rightarrow 0 ^ { + } } \frac { -\sin x } { 2x } = \frac{-1}{2}$.Therefore, the correct answer is $- \frac { 1 } { 2 }$.

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