Limiting roots

Calculus Level 1

Find
$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { \sqrt { x+1 } -1\quad }{ x } }$

This problem is a part of my set The Best of Me

The answer is 0.5.

1 solution

Siddharth G
Jun 2, 2014

Let $x=n-1$
Then
$\lim _{ n-1\rightarrow 0 }{ \frac { \sqrt { (n-1)+1 } -1\quad }{ n-1 } }$
$\Rightarrow \lim _{ n\rightarrow 1 }{ \frac { \sqrt { n } -1\quad }{ n-1 } }$

Since $n-1=(\sqrt { n } -1)(\sqrt { n } +1)$

$\Rightarrow \lim _{ n\rightarrow 1 }{ \frac { \sqrt { n } -1\quad }{ (\sqrt { n } -1)(\sqrt { n } +1)} }$ $\Rightarrow \lim _{ n\rightarrow 1 }{ \frac { 1\quad }{ \sqrt { n } +1 } }$ $=\frac { 1 }{ \sqrt { 1 } +1 }$ $=\frac { 1 }{ 2 }$ = $\boxed{0.5}$

You can do this a different way. One can use L’Hospital’s Rule to solve this. lim {x -> 0} \frac{\sqrt{x+1}-1}{x}=lim {x -> 0} \frac{a}{b} = = \frac{lim {x -> 0}a'}{lim {x -> 0}b'} where a'=\frac{x+1}{2} and b'=1 now we can simply sub in the value zero and find that the limit is 1/2

Matt Schramm - 6 years, 10 months ago

Limiting roots

The answer is 0.5.

1 solution

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