Limit's Dilemma

Calculus Level 3

lim x 0 ( 1 + a x + b x 2 ) 2 x = e 3 \large \lim_{x\to0} (1+ax+bx^2)^\frac 2x = e^3

If the equation above holds true for constants a a and b b , which of the following options must be true?

Notation: R \mathbb R denotes the set of all real values .

a = 1.5 , b R a=1.5, b \in \mathbb R a R , b = 0 a\in \mathbb R, b =0 a = 1.5 , b = 0 a = 1.5, b = 0 a = 0 , b = 0 a = 0, b = 0 a = 1 , b = 1 a = 1, b = 1

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1 solution

Akshay Sharma
May 4, 2016

Since the limit is in the form of 1 \boxed{1^{\infty}}

Now, Generalised formula for 1 1^{\infty} if lim x a f ( x ) = 1 \boxed{\lim_{x\to a}f(x)=1} and lim x a ± \boxed {\lim_{x\to a}\pm \infty} is

lim x 0 ( f ( x ) ) g ( x ) = e lim x 0 g ( x ) ( f ( x ) 1 ) \boxed {\large \lim_{x\to0}(f(x))^{g(x)}=e^{\lim_{x\to0}g(x)(f(x)-1)}}

Considering f ( x ) = 1 + a x + b x 2 f(x)=1+a x+b x^2 and g ( x ) = 2 x g(x)=\frac{2}{x}

Applying above formula lim x 0 ( 1 + a x + b x 2 ) 2 / x = e lim x 0 2 x ( a x + b x 2 ) \large \lim_{x\to0} (1+ax+bx^2)^{2/x} = e^{\lim_{x\to0}\frac{2}{x}(a x +b x^2)}

According to the question , e lim x 0 2 x ( a x + b x 2 ) = e 3 \large e^{\lim_{x\to0}\frac{2}{x}(a x +b x^2)}=e^3

Comparing the exponents, we get

lim x 0 2 x ( a x + b x 2 ) = 3 \large \implies {\lim_{x\to0}\frac{2}{x}(a x +b x^2)}=3

lim x 0 2 a + 2 b x = 3 \large \implies \lim_{x\to0} 2 a+2 b x=3 2 a + 2 b ( 0 ) = 3 \large \implies 2 a+2 b (0)=3

2 a = 3 \large \implies 2 a=3 a = 1.5 , b R \large \implies \boxed {a=1.5,b\in \mathbb R}

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