Find the value of $\lim_{x\to 0} \frac {(1+3x+2x^2)^{1/x} - (1+3x-2x^2)^{1/x}}{x}$

Question

I can apply L'Hopital rule on the very first step itself, but it is quite cumbersome. Is there any alternate method?

My book does imply there is, but it is badly explained.

Answer 1

$$ \begin{align*} &\lim_{x\rightarrow 0} \frac{\left( 1+3x+2x^2 \right) ^{\frac{1}{x}}-\left( 1+3x-2x^2 \right) ^{\frac{1}{x}}}{x} \\ =&\lim_{x\rightarrow 0} \frac{e^{\frac{1}{x}\ln \left( 1+3x+2x^2 \right)}-e^{\frac{1}{x}\ln \left( 1+3x-2x^2 \right)}}{x} \\ =&\lim_{x\rightarrow 0} \frac{e^{\frac{1}{x}\ln \left( 1+3x-2x^2 \right)}\left( e^{\frac{\ln \left( 1+3x+2x^2 \right) -\ln \left( 1+3x-2x^2 \right)}{x}}-1 \right)}{x} \\ =&e^3\lim_{x\rightarrow 0} \frac{\ln \left( \frac{1+3x+2x^2}{1+3x-2x^2} \right)}{x^2} \\ =&e^3\lim_{x\rightarrow 0} \frac{4x^2}{x^2\left( 1+3x-2x^2 \right)} \\ =&e^3\lim_{x\rightarrow 0} \frac{4}{1+3x-2x^2} \\ =&4e^3 \end{align*} $$

Answer 2

Can you figure out the first two terms of the series expansions of $(1 + 3x + 2x^2)^{1/x}$ and $(1 + 3x - 2x^2)^{1/x}$ around $x = 0$? If yes, you are done. It may be helpful to observe that $$(1 + 3x + 2x^2)^{1/x} = e^{\frac{\log(1+3x + 2x^2)}{x}}$$ $$(1 + 3x - 2x^2)^{1/x} = e^{\frac{\log(1+3x - 2x^2)}{x}}$$ For your convenience, I'm telling you that (easily verifiable), $$(1 + 3x + 2x^2)^{1/x} = e^3 - \frac{5e^3x}{2} + \ldots $$ $$(1 + 3x - 2x^2)^{1/x} = e^3 - \frac{13e^3x}{2} + \ldots $$ So, $$\frac{(1 + 3x + 2x^2)^{1/x} - (1 + 3x - 2x^2)^{1/x}}{x} = 4e^3 + \ldots $$ which means $$\lim_{x\to 0}\frac{(1 + 3x + 2x^2)^{1/x} - (1 + 3x - 2x^2)^{1/x}}{x} = 4e^3$$ I have not computed any higher-order terms, because we do not need them - terms with some non-negative power of $x$ go to $0$ as $x\to 0$.

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How to compute terms of a series expansion of $f(x)$ around $x = 0$? $$f(x) = f(0) + \frac{f'(0)x}{1!} + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x}{3!} + \ldots$$ For your question, only the first derivatives at zero are needed, which are easy to compute.