What is the limit as x approaches infinity of sqrt((8x^2-3x)/(16x^2-1)) ?

The limit as x approaches infinity of sqrt((8x^2-3x)/(16x^2-1)) is sqrt(1/2)

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limit as x approaches infinity of sqrt((8x^2-3x)/(16x^2-1))

x \to \infty lim (\frac{8 x ^{2} - 3 x}{16 x ^{2} - 1})

\frac{1}{2}

Decimal Notation

0.70710 \dots

Solution steps

x \to \infty lim (\frac{8 x ^{2} - 3 x}{16 x ^{2} - 1})

Divide by highest denominator power:

\frac{8 - \frac{3}{x}}{16 - \frac{1}{x ^{2}}}

= x \to \infty lim (\frac{8 - \frac{3}{x}}{16 - \frac{1}{x ^{2}}})

x \to a lim [f (x)]^{b} = [x \to a lim f (x)]^{b}

With the exception of indeterminate form

= x \to \infty lim (\frac{8 - \frac{3}{x}}{16 - \frac{1}{x ^{2}}})

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} ( 8 - \frac{3}{x} )}{lim _{x \to \infty} ( 16 - \frac{1}{x ^{2}} )}

x \to \infty lim (8 - \frac{3}{x}) = 8

x \to \infty lim (16 - \frac{1}{x ^{2}}) = 16

= \frac{8}{16}

Cancel the common factor:

8

= \frac{1}{2}

x \to - 2 lim (2 - x)

x \to \infty lim ([3, x])

x \to - 2 lim (7 x + 1)

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

x \to \infty lim (\frac{2 x}{2})

What is the limit as x approaches infinity of sqrt((8x^2-3x)/(16x^2-1)) ?
The limit as x approaches infinity of sqrt((8x^2-3x)/(16x^2-1)) is sqrt(1/2)