Evaluate lim x->0 [√(1+x) -1]/x when I sub in 0 i get 0/0 I am trying to simply the expression or cancel terms but I am having no luck. the answer at the back of the book is 1/2.
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Originally Posted by edwardkiely Evaluate lim x->0 [√(1+x) -1]/x when I sub in 0 i get 0/0 I am trying to simply the expression or cancel terms but I am having no luck. the answer at the back of the book is 1/2. Rationalize the numerator ... $\dfrac{\sqrt{1+x}-1}{x} \cdot \dfrac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$ ... then determine the limit.
when I sub 0 into the simplified version I get 0
Originally Posted by skeeter Rationalize the numerator ... $\dfrac{\sqrt{1+x}-1}{x} \cdot \dfrac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$ ... then determine the limit. $\dfrac{\sqrt{1+x}-1}{x} \cdot \dfrac{\sqrt{1+x}+1}{\sqrt{1+x}+1} = \dfrac{(1+x)-1}{x(\sqrt{1+x}+1} = \dfrac{1}{\sqrt{1+x}+1}$ $\lim \limits_{x\to 0} \dfrac{1}{\sqrt{1+x}+1} = \dfrac{1}{\sqrt{1+0}+1} = \dfrac 1 2$
$\displaystyle \dfrac{\sqrt{1+x}-1}{x} \cdot \dfrac{\sqrt{1+x}+1}{\sqrt{1+x}+1} = \frac{ (1+x)-1}{x\left(\sqrt{1+x}+1\right)} = \frac {x}{x\left(\sqrt{1+x}+1\right)} = \frac {1}{\sqrt{1+x}+1}$