Question

Find the limit. Problem in comments

Answer 1

\[\lim x \rightarrow-\infty (x+\sqrt{x^2+6x})\]

Answer 2

rewrite it
\[x + \sqrt{x} \sqrt{x+6}\]
for infinite limits, added constants are insignificant
\[\sqrt{x+6} = \sqrt{x} , x \rightarrow \pm \infty\]
now limit becomes
\[\lim_{x \rightarrow -\infty} x + x = 2x\]

Answer 3

what do I do after that? They are looking for a number answer

Answer 4

oh wait limit is indeterminate because
\[\sqrt{x^2} = |x|\]
\[\lim_{x \rightarrow -\infty} x + |x| = -\infty + \infty\]

Answer 5

@dumbcow can we multiply both numerator and denominator by x - sqrt(x^2 +6x)?
after simplifying, we get lim = + infinitive only.

Answer 6

hmm yeah i guess, wouldnt that give infty/infy which is still indeterminate ?

Answer 7

\(\lim \limits_{x \rightarrow-\infty} (x+\sqrt{x^2+6x}) \)

\(\lim \limits_{x \rightarrow + \infty} (\sqrt{x^2-6x}-x) \)

multiplying by conjugate and simplifying you get :
\(\lim \limits_{x \rightarrow + \infty} \dfrac{-6x}{\sqrt{x^2-6x}+x} \)

Answer 8

factor out x from the denominator and cancel it with numerator
next take the limit