What is the limit as x approaches 2+of (x-1)/(x^2-4) ?

The limit as x approaches 2+of (x-1)/(x^2-4) is infinity

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limit as x approaches 2+of (x-1)/(x^2-4)

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

\infty

Solution steps

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

Multiply by the conjugate of

x - 1 : \frac{x ^{2} - 1}{x + 1}

= x \to 2 + lim (\frac{\frac{x ^{2} - 1}{x + 1}}{x ^{2} - 4})

Simplify

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

= x \to 2 + lim (\frac{x - 1}{( x + 2 ) ( x - 2 )})

\frac{x - 1}{( x + 2 ) ( x - 2 )} = (x - 1) \frac{1}{( x + 2 ) ( x - 2 )}

= x \to 2 + lim ((x - 1) \frac{1}{( x + 2 ) ( x - 2 )})

x \to a lim [f (x) \cdot g (x)] = x \to a lim f (x) \cdot x \to a lim g (x)

With the exception of indeterminate form

= x \to 2 + lim (x - 1) \cdot x \to 2 + lim (\frac{1}{( x + 2 ) ( x - 2 )})

x \to 2 + lim (x - 1) = 1

x \to 2 + lim (\frac{1}{( x + 2 ) ( x - 2 )}) = \infty

= 1 \cdot \infty

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

t \to 0 lim (\frac{sin ( 1 - cos ( t ) )}{1 - cos ( t )})

x \to 1 lim (2 x - 1)

x \to 0 lim (\frac{( 5 ^{x} - 6 ^{x} )}{x})

x \to - 7 + lim (f (x))

x \to - 9 lim (∣ x + 9 ∣)

What is the limit as x approaches 2+of (x-1)/(x^2-4) ?
The limit as x approaches 2+of (x-1)/(x^2-4) is infinity