# Solve this limit problem

Note by Raja Metronetizen
4 years, 9 months ago

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The trick is to state $$\large (1 + x)^{ \frac {1}{x}}$$ in terms of its Maclaurin Series (up to quadratic powers, because the denominator is a quadratic power as well).

Recall that the Maclaurin Series of $$f(x)$$ is $$\displaystyle \sum_{k=0}^\infty \frac { f^{k} (0) }{k!} x^k$$, where $$f^{k} (0)$$ denote the $$k^{th}$$ derivative of $$f(x)$$ at $$x=0$$, with that we get $$\ln (1 + x) = x - \frac {x^2}{2} + \frac {x^3}{3} - \frac {x^4}{4} + \ldots$$. And apply the quotient rule: if $$f(x) = \frac {u(x)}{v(x)}$$, then $$\large f'(x) = \frac {v(x) u'(x) - u(x) v'(x) }{ [v(x)]^2}$$

Then for small $$x$$, $$\ln (1+x) \approx x - \frac {x^2}{2} + \frac {x^3}{3}$$, $$(*)$$

Let $$\large f(x) = (1 + x)^{\frac {1}{x}}$$, we get $$f(0) = \displaystyle \lim_{x \to 0} (1 + x)^{\frac {1}{x}} = e$$

$$\large \Rightarrow \ln f(x) = \frac {\ln (1+x) }{x}$$

$$\large \Rightarrow \frac { f'(x) }{f(x)} = \frac { \frac {x}{1+x} - \ln (1+x) }{x^2} = \frac {x - (1+x) \space \ln (1+x) }{ x^2 (1 + x) }$$

For small $$x$$, substitute $$(*)$$

$$\large \Rightarrow \LARGE \frac { f'(x) }{f(x)} = \frac {x - (1+x) \space (x - \frac {x^2}{2} + \frac {x^3}{3} ) }{ x^2 (1 + x) }$$

$$\large \Rightarrow \LARGE \frac { f'(x) }{f(x)} = \frac {-2x^2 + x - 3} {6x+6}$$, $$(**)$$

$$\large \Rightarrow f'(0) = f(0) \cdot \frac {-3}{6} = -\frac {e}{2}$$, $$(***)$$

Differentiate $$(**)$$:

$$\large \frac { f(x) f''(x) - [ f'(x) ]^2 }{ [ f(x) ]^2 } = - \frac {x^2 + 2x - 2}{3x^2 + 6x + 3 }$$

$$\large \Rightarrow \frac { f(0) f''(0) - [ f'(0) ]^2 }{ [ f(0) ]^2 } = \frac {2}{3}$$

$$\LARGE\Rightarrow \frac { e \cdot f''(0) - [ - \frac {e}{2} ]^2 }{ e^2 } = \frac {2}{3}$$

$$\large \Rightarrow f''(0) = \frac {11e}{12}$$, $$(****)$$

Therefore, for small $$x$$, $$\large (1 + x)^{ \frac {1}{x} } = f(x) = \displaystyle \sum_{k=0}^\infty \frac { f^{k} (0) }{k!} x^k$$

$$\large \Rightarrow (1 + x)^{ \frac {1}{x} } = e - \frac {ex}{2} + \frac {11ex^2}{24} + O(x^3)$$

Now we can evaluate the limit

$$\large \displaystyle \lim_{x \to 0} \frac { (1+x)^{\frac {1}{x} } - e + \frac {ex}{2} } {x^2}$$

$$\large = \displaystyle \lim_{x \to 0} \frac { 2(1+x)^{\frac {1}{x} } - 2e + ex } {2x^2}$$

$$\large = \displaystyle \lim_{x \to 0} \frac { 2(e - \frac {ex}{2} + \frac {11ex^2}{24} + O(x^3) ) - 2e + ex } {2x^2}$$

$$\large = \displaystyle \lim_{x \to 0} \frac { \frac { 11ex^2}{12} + O(x^3) } {2x^2}$$

$$\large = \displaystyle \lim_{x \to 0} \frac { \frac { 11e}{12} + O(x) } {2}$$

$$\large = \frac {11e}{24}$$

- 4 years, 9 months ago

How you format these equations in solutions? Can we use latex?

- 4 years, 9 months ago

Amazing! Great job Pi Han!

- 4 years, 9 months ago

Thanks you a lot...Great job it is....U r seemingly the master hand of limit problems...Thanks.

- 4 years, 9 months ago

Here is another method :

$$\displaystyle \lim_{x \to 0} \frac{(1 + x)^{\frac{1}{x}} - e + \frac{ex}{2}}{x^2}$$

= $$\displaystyle \lim_{x \to 0} \frac{e^{\frac{ln(1 + x)}{x}} - e + \frac{ex}{2}}{x^2}$$

= $$\displaystyle \lim_{x \to 0} \frac{e \Big(e^{\frac{ln(1 + x)}{x} - 1} - 1 + \frac{x}{2} \Big)}{x^2}$$

= $$\displaystyle \lim_{x \to 0} \frac{e \Big(e^{\frac{\big(x - \frac{x^2}{2} + \frac{x^3}{3} - \dots \big)}{x} - 1} - 1 + \frac{x}{2} \Big)}{x^2}$$

= $$\displaystyle \lim_{x \to 0} \frac{e \Big(e^{(-\frac{x}{2} + \frac{x^2}{3} - \dots )} - 1 + \frac{x}{2} \Big)}{x^2}$$

= $$\displaystyle \lim_{x \to 0} \frac{e \Big( 1 + (-\frac{x}{2} + \frac{x^2}{3} - \dots ) + \frac{{(-\frac{x}{2} + \frac{x^2}{3} - \dots )}^2}{2} + \dots - 1 + \frac{x}{2} \Big)}{x^2}$$

$$\displaystyle \lim_{x \to 0} \frac{e(\frac{x^2}{3} + \frac{x^2}{8} + \dots)}{x^2}$$ = $$\boxed{\frac{11e}{24}}$$

- 4 years, 9 months ago

write (1+x)^(1/x)=e^(1/x)log(1+x) and expand carefully by neglecting all the terms greater than x^2, you can get it

- 4 years, 9 months ago