Question Number 100920 by ajfour last updated on 29/Jun/20

Find limit      lim_(x→+∞) x((√(x^2 +1))−x)   and     lim_(x→−∞) x((√(x^2 +1))−x)  .

Commented bybramlex last updated on 29/Jun/20

lim_(x→+∞)  x(x(√(1+(1/x^2 )))−x)   set (1/x) = z ,z→0  lim_(z→0)  (1/z)((1/z)(√(1+z^2 ))−(1/z)) =  lim_(z→0)  (((√(1+z^2 )) −1)/z^2 ) = lim_(z→0)  (((1+(z^2 /2))−1)/z^2 )  = (1/2) ★

Commented bybramlex last updated on 29/Jun/20

lim_(x→−∞)  x((√(x^2 +1))−x)=  lim_(x→−∞) x(−x(√(1+(1/x^2 )))−x)=  lim_(x→−∞) x(−x(√(1+((1/(−x)))^2 ))−x)=  lim_(x→−∞) −x^2 ((√(1+(−(1/x))^2 ))+1)  set (1/(−x)) = q , q→0  lim_(q→0)  −((1/q))^2 ((√(1+q^2 ))+1)=  −lim_(q→0)  (((√(1+q^2 ))+1)/q^2 ) = −∞ .★

Answered by mathmax by abdo last updated on 29/Jun/20

for x>0 we have x((√(x^2 +1))−x) =x(x(√(1+(1/x^2 )))−x)∼x(x(1+(1/(2x^2 )))−x)  =x( (1/(2x))) =(1/2) ⇒lim_(x→+∞)  x((√(1+x^2 ))−x) =(1/2)  for x<0  we have x((√(x^2 +1))−x) =x{ −x(√(1+(1/x^2 )))−x}  ∼x{−x(1+(1/(2x^2 )))−x} =x{−2x−(1/(2x))} =−2x^2 −(x/2) →−∞

Commented byajfour last updated on 29/Jun/20

correct Sir,  i mean lim_(x→+∞) f(x)=(1/2)  and  lim_(x→−∞) f(x)=−∞ . Thanks.

Commented bymathmax by abdo last updated on 29/Jun/20

you are welcome