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

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

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

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

\frac{7}{2}

Decimal

3.5

Solution steps

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

Multiply by the conjugate of

x^{2} + 7 x + 2 - x : \frac{7 x + 2}{x ^{2} + 7 x + 2 + x}

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

Divide by highest denominator power:

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

= x \to \infty lim \frac{7 + \frac{2}{x}}{1 + \frac{7}{x} + \frac{2}{x ^{2}} + 1}

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

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

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

= \frac{7}{2}

x \to 0 + lim (ln (x) sin^{2} (x))

x \to 0 lim (3 x^{2})

x \to \infty lim (\frac{2 x ^{3} - 5 x ^{2} + 4 x - 345}{6 x ^{3} + 2 x})

x \to 0 lim (4 + e^{\frac{1}{x}})

x \to 90 o lim (sin (x))

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