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\[\large \lim_{x\to1} \frac{x+x^2+x^3+\ldots +x^n-n}{\sqrt x-1} = \ ? \]

Note by A K
2 years ago

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Without L'Hôpital's rule

Rationalize the denominator by multiplying the expression with \(\frac{\sqrt{x} + 1} {\sqrt{x} + 1}\). Distribute the \(n\) and rewrite the expression as

\[ \lim _{ x \to 1 } \frac{ (x - 1) + (x^{2} - 1) + (x^{3} - 1) + \cdots + (x^{n} - 1) }{x -1} \cdot (\sqrt{x} + 1)\]

\[ = \lim _{ x \to 1 } \left( \frac{ x - 1 } { x -1 } + \frac{ x^{2} - 1 } { x -1 } + \frac{ x^{3} - 1 } { x -1 } + \cdots + \frac{ x^{n} - 1 } { x -1 }\right) \cdot (\sqrt{x} + 1)\]

\[ = \lim _{ x \to 1 } \left( (1) + (1 + x) + (1 + x + x^{2}) + (1 + x + x^{2} + x^{3}) + \cdots + (1 + x + x^{2} + x^{3} + \cdots + x^{n-1}) \right) \cdot (\sqrt{x} + 1)\]

Now solve by substitution:

\[ = ( 1 + 2 + 3 + \cdots + n) \cdot (1 + 1) = \boxed{n(n+1)}\] Pranshu Gaba · 2 years ago

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@Pranshu Gaba Amazing solution! :) John Frank · 1 year, 8 months ago

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n(n+1) Jahid Rafi · 1 year, 3 months ago

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n(n+1) Akshay Singh Sengar · 2 years ago

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@Akshay Singh Sengar Did you use L.Hospital's Rule ? A K · 2 years ago

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n(n+1) May be thats the answer. If it is correct please notify. Mukul Sharma · 2 years ago

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@Mukul Sharma Did you use L.Hospital's Rule ? A K · 2 years ago

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n(n+1) Vincent Miller Moral · 2 years ago

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@Vincent Miller Moral Did you use L.Hospital's Rule ? A K · 2 years ago

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@A K Yes. However I cannot post my solution. My Latex skills s*cks. Sorry. Vincent Miller Moral · 2 years ago

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\(n^{2}+n\)??? Aditya Kumar · 2 years ago

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@Aditya Kumar Did you use L.Hospital's Rule ? A K · 2 years ago

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@A K Yes. Can u prove the L.hospital rule? Aditya Kumar · 2 years ago

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Since this is an indeterminate form of the \(\dfrac{0}{0}\) form, we can simply use the L.Hospital's Rule to evaluate the given limit, which is as you follow:

\[\begin{array}{} & \lim_{x\to1} \dfrac{x+x^2+x^3+\ldots +x^n-n}{\sqrt x-1} \\ & = \lim_{x \to 1} \dfrac{1+2x+3x^2+\ldots +nx^{n-1}}{\dfrac{1}{2 \sqrt x}} \\ & = n(n+1) \end{array} \] Sandeep Bhardwaj · 2 years ago

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@Sandeep Bhardwaj There is an answer without using L.Hospital's Rule ,....can u find it ? A K · 2 years ago

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@A K There will be no difference in the answer whether you solve it using L.Hospital's rule or not. But yeah, there obviously exists a way to evaluate it without using L.Hospital's rule, as Pranshu did. And notice that the answer is still the same. Sandeep Bhardwaj · 2 years ago

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@Sandeep Bhardwaj We could just use L'Hopital's rule but sometimes we can search about another answer ...I was looking for an approach that does not use this rule and I am sorry I didn't mention , but your solution still elegant :)) A K · 2 years ago

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