Verified answer

= lim(x→∞)-x/(x+√(x²+x)) = -1/2

yes, it's

lim ( x - √(x² + x ) ) ===> large in charge

x--->∞

lim ( x - √(x²)

x--->∞

lim ( x - x ) = ∞ - ∞ ( not determined )

x--->∞

I would say let's start conjugating as the following:

( x - √(x² + x ) ) * [ ( x + √(x² + x ) ) / ( x + √(x² + x ) ) ]

[ x^2 - (x² + x ) / ( x + √(x² + x ) ) ]

[ (x^2 - x² - x ) / ( x + √(x² + x ) ) ]

[ - x / ( x + √(x² + x ) ) ] ====> algebra technique

[ - x / x( 1 + √(x² + x )/x ) ]

[ - 1 / ( 1 + √( (x² + x )/x^2 ) ]

[ - 1 / ( 1 + √( (x²/x^2 + x/x^2 ) ]

[ - 1 / ( 1 + √( (1 + 1/x ) ]

lim [ - 1 / ( 1 + √( (1 + 1/x ) ] =

x--->∞

[ - 1 / ( 1 + √( (1 + 1/∞ ) ]

[ - 1 / ( 1 + √( (1 + 0 ) ]

[ - 1 / ( 1 + √(1) ]

[ - 1 / ( 1 + 1 ]

(- 1/2)

============

the next problem is not similar.

( √(x²+1) - √(x) ) ====> conjugate

( √(x²+1) - √(x) ) * ( √(x²+1) + √(x) ) / ( √(x²+1) + √(x) )

( (x²+1) - x ) / ( √(x²+1) + √(x) )

( x²+1 - x ) / ( √(x²+1) + √(x) ) ====> algebra technique : factor out the √(x)

√(x) ( x^(3/2) + 1/√(x) - √(x) ) / √(x)( √(x²+1)/√(x) + 1 )

( x^(3/2) + 1/√(x) - √(x) ) / ( √( (x²+1)/x ) + 1 )

( x^(3/2) + 1/√(x) - √(x) ) / ( √( (x²/x+1/x ) + 1 )

( x^(3/2) + 1/√(x) - √(x) ) / ( √(x + x^-1 ) + 1 )

lim ( x^(3/2) + 1/√(x) - √(x) ) / ( √(x + x^-1 ) + 1 ) ===> large in charge

x--->∞

lim ( x^(3/2) ) / √(x)

x--->∞

lim x^3 = ∞^3 = ∞

x--->∞