limx→−∞1−ex2−x1+ex2−x
I solved this limit problem by applying L'Hôpital's rule and I got −1.
Question: how to solve this limit without L'Hopital rule and Taylor series?
Answer
hint:
Divide Numerator and Denominator by ex2−x
limx→−∞1−ex2−x1+ex2−x=limx→−∞1ex2−x−11ex2−x+1
Now, limx→−∞1ex2−x=0, since limx→−∞x2−x=+∞ and limx→∞ex=+∞
⟹ Rqrd. limit=−1
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