Question

Find the Limit
X----> -infinity

Answer 1

\[x + \sqrt{x^2 + 3}\]

Answer 2

next i get
\[\frac{ x^2 - x^2 +3 }{ x - \sqrt{x^2 +3} }\]

Answer 3

equals negative infinity?

Answer 4

Looks good,
divide top and bottom by x

Answer 5

does the limit equal zero?

Answer 6

i get 3/2(-infinity)

Answer 7

\[(x + \sqrt{x^2 + 3})(x - \sqrt{x^2 + 3})=x^2-(x^2+3)=x^2-x^2-3=-3\]

Answer 8

not that it will change your final answer (because it wont). 0 is the answer

Answer 9

\[\large \begin{align} \\ \lim\limits_{x\to -\infty}~ x + \sqrt{x^2 + 3}
& =\lim\limits_{x\to -\infty} \dfrac{-3 }{x - \sqrt{x^2 + 3}} \\~\\
& =\lim\limits_{y\to \infty} ~\dfrac{-3 }{-y - \sqrt{y^2 + 3}} \\~\\
&= \lim\limits_{1/y\to 0}~ \dfrac{-3/y }{-1 - \sqrt{1 + 3/y^2}} \\~\\ &= \dfrac{0}{-2}=0\end{align}\]