limit help

[orir]

Im sorry i dont use math here. It's because i cant find where i do this...

Can anyone please explain me how do i find the limit of f(x) when x approaches (-infinity)?
f(x)=sqrt(x^2+1)-x

 

[DrPhil]

Im sorry i dont use math here. It's because i cant find where i do this...

Can anyone please explain me how do i find the limit of f(x) when x approaches (-infinity)?
f(x)=sqrt(x^2+1)-x

Please show your work!

Are you sure you have signs right?

When x is negative, both terms are positive, so . . .

When x is large and positive, use the approximation \(\displaystyle \sqrt{1 + \delta}\ \approx\ 1 + \delta /2\)

 

[HallsofIvy]

Im sorry i dont use math here. It's because i cant find where i do this...

Can anyone please explain me how do i find the limit of f(x) when x approaches (-infinity)?
f(x)=sqrt(x^2+1)-x

Generally speaking for something like this, the best thing to do is to multiply by the fraction (sqrt(x^2+ 1)+ x)/(sqrt(x^2+1)+ x) to "rationalize" the numerator, giving (x^2+ 1- x^2)/(sqrt(x^2+1)+ x)= 1/(sqrt(x^2+1)+ x. Now, as Dr Phil pointed out, for x going to negative infinity, say x= -yfor some positive y, for y very very large, that will be close to 1/(y- y).

 

[lookagain]

Generally speaking for something like this, the best thing to do is to multiply by the fraction (sqrt(x^2+ 1)+ x)/(sqrt(x^2+1)+ x) to "rationalize" the numerator,

giving (x^2+ 1- x^2)/(sqrt(x^2+1)+ x)= 1/(sqrt(x^2+1)+ x).
You missed a right parenthesis.

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