What is the limit as x approaches-infinity of ((sqrt(4x^2+1)))/((2-7x)) ?

The limit as x approaches-infinity of ((sqrt(4x^2+1)))/((2-7x)) is 2/7

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limit as x approaches-infinity of ((sqrt(4x^2+1)))/((2-7x))

x \to - \infty lim (\frac{( 4 x ^{2} + 1 )}{( 2 - 7 x )})

\frac{2}{7}

Decimal

0.28571 \dots

Solution steps

x \to - \infty lim (\frac{4 x ^{2} + 1}{2 - 7 x})

Divide by highest denominator power:

\frac{\frac{4 x ^{2} + 1}{x}}{\frac{2}{x} - 7}

= x \to - \infty lim (\frac{\frac{4 x ^{2} + 1}{x}}{\frac{2}{x} - 7})

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} ( \frac{4 x ^{2} + 1}{x} )}{lim _{x \to - \infty} ( \frac{2}{x} - 7 )}

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

x \to - \infty lim (\frac{2}{x} - 7) = - 7

= \frac{- 2}{- 7}

Simplify

= \frac{2}{7}

x y^{^{'}} + 4 y = - 25 x sin (x^{5})

\frac{d}{d x} (\frac{3}{7 x ^{4} - 5 x ^{2} + 1})

\frac{d y}{d x} = tan (4 x)

\int 16 - t^{2} d t

\frac{\partial}{\partial y} (1 - x - y)

What is the limit as x approaches-infinity of ((sqrt(4x^2+1)))/((2-7x)) ?
The limit as x approaches-infinity of ((sqrt(4x^2+1)))/((2-7x)) is 2/7