Question

If $\lim_{x \to 0} \left(1+ax+bx^2\right)^{\frac{2}{x}}\:=\:e^3$ , find a & b

Answer 1

Answer:

Given:

$\lim_{x \to 0}\: (1 + ax + bx^2 )^{2/x} = e^3$

On substituting x = 0, we get the form: $1^\infty$

This indeterminate form is usually evaluated as:

$\lim_{x \to 0}\: (f(x))^{g(x)} = 1^\infty \:\: \implies\:\: \lim_{x \to 0}\:\:e^{ g(x) [ f(x) - 1 ]}$

Converting it to the form mentioned above we get:

$\lim_{x \to 0}\:\: e^{ (2/x)\:[ 1 + ax + bx^2 - 1 ]} = e^3\\\\\\\text{Multiplying the powers we get:}\\\\\\\implies \dfrac{ 2}{x} ( ax ) + \dfrac{2}{x} ( bx^2)\:\; \implies \boxed{2a + 2bx}$

$\text{Hence the new form would be:}\\\\\\\implies \lim_{x \to 0} e^{ (2a + 2bx )} = e^3\\\\\\\text{Applying limts we get:}\\\\\\\implies e^{(2a + 2b(0))} = e^3\\\\\\\implies e^{2a} = e^3\\\\\\\implies 2a = 3\:\: \implies \boxed{a = \dfrac{2}{3}}$

Since we get 'b' to be independent, 'b' can take all real values.

Answer 2

$\underbrace{ \boxed{ \underline{ \blue{ \underline{\tt \red{Solution }}}}}}$

$\green{ \tt we \: have} \\ \tt \lim _{x \rightarrow0}(1 + ax + {bx}^{2} {)}^{ \frac{2}{x} } = {e}^{3} \: \: ( {1}^{\infin} form) \\ \tt \implies ( \lim _{x \rightarrow0}(1 + ax + {bx}^{2} {)}^{ \frac{1}{ax + {bx}^{2} } } {)}^{ \frac{2(ax + {bx}^{2}) }{x} } = {e}^{3} \\ \tt\implies {e}^{2a} = {e}^{ 3} \\ \tt \implies 2a = 3 \\ \tt \implies a = \frac{3}{2} \: and \: b ∈R. \\ \\ \\ \boxed{ \tt a = \frac{3}{2} \: and \: b ∈R }$