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

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

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

1

Solution steps

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

Simplify

\frac{( x + 1 ) ^{2} + 1}{x ^{2} + 1} : \frac{x ^{2} + 2 x + 2}{x ^{2} + 1}

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

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{x ^{2} + 2 x + 2}{x ^{2} + 1})

Divide by highest denominator power:

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

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

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

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

= \frac{1}{1}

Simplify

= 1

x \to 1 lim (\frac{10 - x - 8 + x}{x ^{3} - 1})

x \to - 3 + lim (- \frac{sin ( c - 3 x )}{e ^{5 x + 15} - 1})

x \to 0 + lim (\frac{e ^{x} - 1 - x}{x ^{2}})

x \to \infty lim (\frac{6 x ^{6} + x}{7 x ^{5} - 2 x})

x \to \infty + lim (\frac{( e ^{x} )}{x})

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