How do you evaluate the limit #sqrt(x^2+1)-2x# as x approaches #oo#?

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Answer 1 · 2016-12-13T19:07:35

See below.

Change the way the expression is written.

If we try to find the limit as written, we get the indeterminate form #oo-oo#.

We could change this to a fraction using #(sqrt(x^2+1)+2x)/(sqrt(x^2+1)+2x)#, but perhaps the following is simpler.

For #x !`= 0#, we have

#sqrtx^2+1) - 2x = sqrt(x^2(1+1/x^2))-2x#

# = sqrt(x^2)sqrt(1+1/x^2)-2x#

For positive #x#, #sqrt(x^2) = x# so

#lim_(xrarroo) (sqrt(x^2+1) - 2x) = lim_(xrarroo)(xsqrt(1+1/x^2)-2x)#

# = lim_(xrarroo)(x(sqrt(1+1/x^2) - 2))#

Which has the form #oo * (1-2)#

Therefore,

#lim_(xrarroo) (sqrt(x^2+1) - 2x) = -oo#