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What is the limit of #(x + (sqrt(x^2 + 2x))# as x goes to negative infinity?

Answer 1

Refer to explanation

Divide and multiply with conjugate #x-sqrt(x^2+2x)#

Hence we have

#lim_(x->-oo)((x+sqrt(x^2+2x))*(x-sqrt(x^2+2x)))/(x-sqrt(x^2+2x))= lim_(x->-oo)(x^2-(x^2+2x))/((x-sqrt(x^2+2x)))= lim_(x->-oo)(2x)/(x(sqrt(1+2/x)-1))= lim_(x->-oo)2/(sqrt(1+2/x)-1)= lim_(x->-oo) (-2)/(1+sqrt(1+2/x))=-1#

Remarks

Notice that since #x# is tending to #-oo# then #x=-sqrt(x^2)#. So when you let #x# enter the square root in the denominator the minus that was originally there becomes a plus due to cancellation.

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Answer 2

Charles Anctil

The limit of (x + (sqrt(x^2 + 2x)) as x goes to negative infinity is negative infinity.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.