What is the limit as x approaches infinity of ((sqrt(2x^2+1)))/(3x-5) ?

The limit as x approaches infinity of ((sqrt(2x^2+1)))/(3x-5) is (sqrt(2))/3

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limit as x approaches infinity of ((sqrt(2x^2+1)))/(3x-5)

x \to \infty lim (\frac{( 2 x ^{2} + 1 )}{3 x - 5})

\frac{2}{3}

Solution steps

x \to \infty lim (\frac{2 x ^{2} + 1}{3 x - 5})

Divide by highest denominator power:

\frac{2 + \frac{1}{x ^{2}}}{3 - \frac{5}{x}}

= x \to \infty lim \frac{2 + \frac{1}{x ^{2}}}{3 - \frac{5}{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} ( 2 + \frac{1}{x ^{2}} )}{lim _{x \to \infty} ( 3 - \frac{5}{x} )}

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

x \to \infty lim (3 - \frac{5}{x}) = 3

= \frac{2}{3}

y^{^{''}} - 2 y^{^{'}} = 2 t

x \to 3 + lim (\frac{x ^{2} + x}{x - 3})

cos (t + \frac{π}{4})

\int (3 y + 7) d y

\frac{5}{s + 2}

What is the limit as x approaches infinity of ((sqrt(2x^2+1)))/(3x-5) ?
The limit as x approaches infinity of ((sqrt(2x^2+1)))/(3x-5) is (sqrt(2))/3