What is the limit as x approaches 2 of (x^2-4)^{-1}sin(x-2) ?

The limit as x approaches 2 of (x^2-4)^{-1}sin(x-2) is 1/4

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limit as x approaches 2 of (x^2-4)^{-1}sin(x-2)

x \to 2 lim ((x^{2} - 4)^{- 1} sin (x - 2))

\frac{1}{4}

Decimal Notation

0.25

Solution steps

x \to 2 lim ((x^{2} - 4)^{- 1} sin (x - 2))

Simplify

(x^{2} - 4)^{- 1} sin (x - 2) : \frac{sin ( x - 2 )}{x ^{2} - 4}

= x \to 2 lim (\frac{sin ( x - 2 )}{x ^{2} - 4})

Apply L'Hopital's Rule

: x \to 2 lim (\frac{cos ( x - 2 )}{2 x})

= x \to 2 lim (\frac{cos ( x - 2 )}{2 x})

Plug in the value

x = 2

= \frac{cos ( 2 - 2 )}{2 \cdot 2}

Simplify

\frac{cos ( 2 - 2 )}{2 \cdot 2} : \frac{1}{4}

= \frac{1}{4}

y^{^{''}} + 4 y = 6 sin (3 t)

\int 6 sin^{4} (x) cos (x) d x

x^{3} d x + y^{3} d y = 0

\int sin^{5} (t) cos^{4} (t) d t

f^{^{'}} (x) = \frac{1 - x}{x ^{2}}

What is the limit as x approaches 2 of (x^2-4)^{-1}sin(x-2) ?
The limit as x approaches 2 of (x^2-4)^{-1}sin(x-2) is 1/4