J's answer to Bryan's International Baccalaureatte Further Maths HL Singapore question.
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Answered in comments.
Date Posted:
3 years ago
thanks alot !
2b)
lim (e^(-1/x²) / x²)
x→ 0
= lim (1/x² / e^(1/x²) )
x→ 0
= lim (d/dx (1/x²) / d/dx e^(1/x²) )
x→ 0
= lim -2/x³ / (-2/x³ · e^(1/x²) )
x→ 0
= lim 1 / e^(1/x²)
x→ 0
= lim e^(-1/x²)
x → 0
= e^(lim -1/x²)
x→0
= e^(-lim 1/x²)
x→0
= 1/e^(lim 1/x²)
x → 0
= 1/e^∞
= 0
lim (e^(-1/x²) / x²)
x→ 0
= lim (1/x² / e^(1/x²) )
x→ 0
= lim (d/dx (1/x²) / d/dx e^(1/x²) )
x→ 0
= lim -2/x³ / (-2/x³ · e^(1/x²) )
x→ 0
= lim 1 / e^(1/x²)
x→ 0
= lim e^(-1/x²)
x → 0
= e^(lim -1/x²)
x→0
= e^(-lim 1/x²)
x→0
= 1/e^(lim 1/x²)
x → 0
= 1/e^∞
= 0
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