Why is the following limit equal to $1/2$. I get undefined. :-(
$$\lim_{x\to 0}\frac{(x+1)^{1/2}+1}x$$
this should $= 1/2$.
When I multiply the top and bottom by $(x+1)^{1/2} - 1$, I end up with this...
On the numerator: $x+1-1$
On the denominator: $x\Big((x+1)^{1/2} -1\Big)$
This then gets
On the numerator: $1$
On the denominator: $(x+1)^{1/2} -1$
But when as $x\to 0$, the denominator still has a zero! Could someone please help?
$\endgroup$ 62 Answers
$\begingroup$Check the numerator again: you end up with $x$: $\;\;x + 1 - 1 = x$. That term then cancels with the factor of $x$ in the denominator.
$$\frac{(x+1)^{1/2}+1}x\cdot \dfrac{(x+ 1)^{1/2} - 1}{(x+ 1)^{1/2} - 1} = \dfrac {x + 1 - 1}{x(x+ 1)^{1/2} - 1)} = \dfrac{1}{(x + 1)^{1/2}} $$
And now we conclude that the limit of $$\lim_{x \to 0} \dfrac{1}{(x + 1)^{1/2}-1} \to \frac {1}{0}$$
That is, as $x \to 0$, the denominator get's incredibly small, so small that the limit blows up to infinity. Recall that a we don't actually have division by $0$; that is, we are not evaluating the function at zero. What we have is the limit as $x \to 0$, as $x$ gets very very close to zero, but not equal to zero. That is: $$\lim_{x \to 0} \dfrac{1}{(x + 1)^{1/2}-1} = +\infty$$
$\endgroup$ 4 $\begingroup$If you meant $\lim_{x\to0}\frac{\sqrt{x+1}-1}{x}$ then:
$\frac{\sqrt{x+1}-1}{x}=\frac{\sqrt{x+1}-1}{x}\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}=\frac{1}{\sqrt{x+1}+1}$
As $x$ goes to $0$ this goes $\frac{1}{2}$.
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