How do you find #\lim _ { x \rightarrow \infty } \sqrt { x ^ { 2} + 3x } - x#?

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Answer 1 · 2017-07-25T13:59:33

# 3/2.#

#"The Reqd. Limit="lim_(x to oo)(sqrt(x^2+3x)-x),#

#=lim_(x to oo)(sqrt(x^2+3x)-x){sqrt(x^2+3x)+x}/{sqrt(x^2+3x)+x},#

#=lim_(x to oo){(x^2+3x)-x^2}/{sqrt(x^2+3x)+x},#

#=lim_(x to oo) (3x)/{sqrt(x^2(1+3/x))+x},#

#=lim_(x to oo) (3x)/{xsqrt(1+3/x)+x},#

#=lim_(x to oo) (3cancelx)/{cancelx(sqrt(1+3/x)+1)}.#

Knowing that, as # x to oo, 1/x to 0,# we have,

#"The Reqd. Limit="3/(sqrt(1+0)+1)=3/2.#