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

@Aditya Kumar
–
Did you use L.Hospital's Rule ?
–
A K
·
1 year, 6 months ago

Log in to reply

@A K
–
Yes. Can u prove the L.hospital rule?
–
Aditya Kumar
·
1 year, 6 months ago

Log in to reply

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:

@Sandeep Bhardwaj
–
There is an answer without using L.Hospital's Rule ,....can u find it ?
–
A K
·
1 year, 6 months ago

Log in to reply

@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
·
1 year, 6 months ago

Log in to reply

@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
·
1 year, 6 months ago

## Comments

Sort by:

TopNewestWithout 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 · 1 year, 6 months ago

Log in to reply

– John Frank · 1 year, 2 months ago

Amazing solution! :)Log in to reply

n(n+1) – Jahid Rafi · 9 months, 1 week ago

Log in to reply

n(n+1) – Akshay Singh Sengar · 1 year, 6 months ago

Log in to reply

– A K · 1 year, 6 months ago

Did you use L.Hospital's Rule ?Log in to reply

n(n+1) May be thats the answer. If it is correct please notify. – Mukul Sharma · 1 year, 6 months ago

Log in to reply

– A K · 1 year, 6 months ago

Did you use L.Hospital's Rule ?Log in to reply

n(n+1) – Vincent Miller Moral · 1 year, 6 months ago

Log in to reply

– A K · 1 year, 6 months ago

Did you use L.Hospital's Rule ?Log in to reply

– Vincent Miller Moral · 1 year, 6 months ago

Yes. However I cannot post my solution. My Latex skills s*cks. Sorry.Log in to reply

\(n^{2}+n\)??? – Aditya Kumar · 1 year, 6 months ago

Log in to reply

– A K · 1 year, 6 months ago

Did you use L.Hospital's Rule ?Log in to reply

– Aditya Kumar · 1 year, 6 months ago

Yes. Can u prove the L.hospital rule?Log in to reply

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 · 1 year, 6 months ago

Log in to reply

– A K · 1 year, 6 months ago

There is an answer without using L.Hospital's Rule ,....can u find it ?Log in to reply

– Sandeep Bhardwaj · 1 year, 6 months ago

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.Log in to reply

– A K · 1 year, 6 months ago

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 :))Log in to reply