So Close To One

Calculus Level 1

$\large \lim_{x\to1} \dfrac{x^x - 1}{x-1} = \, ?$

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2 solutions

Tay Yong Qiang
Jul 23, 2016

By L' Hôpital's Rule, $\because x^x-1=x-1=0,$ when $x=1$ ,

$\therefore\displaystyle\lim_{x\rightarrow 1} \frac{x^x-1}{x-1}=\lim_{x\rightarrow 1}\frac{x^x(\ln x+1)}{1}=(1^1)(\ln 1 +1)= 1$

Is there wiki for that?

Mr Yovan - 4 years, 10 months ago

Yes - see here: L'Hopital's Rule .

Eli Ross Staff - 4 years, 10 months ago

That's what I did.

Bettie Wallace - 4 years, 7 months ago

Abhishek Kulkarni
Aug 2, 2016

If F(x)= x^n-a^n/x-a then lim x->m f(x)=n(a^n-1). Where; a,n,m are real numbers. Hence, lim x->1 =n(a^n-1) =x(1^x-1) =x But x->1..........thus limit is 1.

I suppose this valid when n is constant and not variable..

Puneet Pinku - 4 years, 8 months ago

This is wrong. In this case, your "n" is a number independent to "x", unlike my question.

Pi Han Goh - 3 years, 4 months ago

So Close To One

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