Evaluate : limx→ −∞(x+√x2+2x)
I've tried some basic algebraic manipulation to get it into a form where I can apply L'Hopital's Rule, but it's still going to be indeterminate form.
This is what I've done so far
limx→ −∞(x+√x2+2x)=limx→ −∞(x+√x2+2x)(x−√x2+2xx−√x2+2x)=limx→ −∞(x2−(x2+2x)x−√x2+2x)=limx→ −∞(−2xx−√x2+2x)
And that's as far as I've gotten. I've tried applying L'Hopitals Rule, but it still results in an indeterminate form.
Plugging it into WolframAlpha shows that the correct answer is −1
Any suggestions on what to do next?
Answer
limx→−∞(−2xx−√x2+2x)=limx→−∞(−2xx−√x2(1+2x))=limx→−∞−2xx−|x|√1+2x==limx→−∞−2xx+x√1+2x=limx→−∞−2xx(1+√1+2x)=−1
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