What is the limit as x approaches infinity of (sqrt(x^2-9))/(2x-6) ?

The limit as x approaches infinity of (sqrt(x^2-9))/(2x-6) is 1/2

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limit as x approaches infinity of (sqrt(x^2-9))/(2x-6)

x \to \infty lim (\frac{x ^{2} - 9}{2 x - 6})

\frac{1}{2}

Decimal

0.5

Solution steps

x \to \infty lim (\frac{x ^{2} - 9}{2 x - 6})

Divide by highest denominator power:

\frac{1 - \frac{9}{x ^{2}}}{2 - \frac{6}{x}}

= x \to \infty lim \frac{1 - \frac{9}{x ^{2}}}{2 - \frac{6}{x}}

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

= \frac{lim _{x \to \infty} ( 1 - \frac{9}{x ^{2}} )}{lim _{x \to \infty} ( 2 - \frac{6}{x} )}

x \to \infty lim (1 - \frac{9}{x ^{2}}) = 1

x \to \infty lim (2 - \frac{6}{x}) = 2

= \frac{1}{2}

y^{^{''}} - 8 y^{^{'}} + 16 y = e^{4 x}

\frac{d}{d x} (4 x^{2} + 6 x)

\frac{\partial}{\partial s} (s^{6} + t^{6})

\frac{df}{d x} + 2 f = f x^{3}

n = 0 \sum \infty (\frac{1}{3})^{n}

What is the limit as x approaches infinity of (sqrt(x^2-9))/(2x-6) ?
The limit as x approaches infinity of (sqrt(x^2-9))/(2x-6) is 1/2