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Solve this limit problem

Please prove this limit problem and please solve it step by step...Thanks in advance...

Note by Raja Metronetizen
3 years, 10 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

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

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

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

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

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

\( \large = \frac {11e}{24} \) Pi Han Goh · 3 years, 10 months ago

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@Pi Han Goh How you format these equations in solutions? Can we use latex? Indulal Gopal · 3 years, 10 months ago

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@Pi Han Goh Amazing! Great job Pi Han! Pranav Arora · 3 years, 10 months ago

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@Pi Han Goh Thanks you a lot...Great job it is....U r seemingly the master hand of limit problems...Thanks. Raja Metronetizen · 3 years, 10 months ago

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Here is another method :

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

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

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

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

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

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

\(\displaystyle \lim_{x \to 0} \frac{e(\frac{x^2}{3} + \frac{x^2}{8} + \dots)}{x^2} \) = \(\boxed{\frac{11e}{24}}\) Jatin Yadav · 3 years, 10 months ago

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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 Ravi Teja · 3 years, 10 months ago

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