Evaluate the limit, if it exists
In Exercises 5-34, evaluate the limit, if it exists. If not, determine whether the one side limit exists (finite or infinite).
26. $$\lim_{x\to 0^+}\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x-1}}\right)$$
So I'm trying to get $x$ out of the denominator, so I tried combining:
$\require{cancel}$
$$\frac{\sqrt{x}}{\sqrt{x}}\cdot \frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x-1}} \cdot \frac{\sqrt{x-1}}{\sqrt{x-1}}$$
$$\frac{\sqrt{x}-\sqrt{x-1}}{-x^2}$$
But I can't go any further.
$\endgroup$ 41 Answer
$\begingroup$The limit does not exist. The function is only defined for values of $x>1$.
However, consider:
$$\lim_{x\to 1^+}\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x-1}}\right) = -\infty$$
Proof:
Notice that as $x\to 1^+$, the left term $\to 1$ while the right term increases without bound (that is, the right term $\to\infty$).
$$\lim_{x\to1^+}\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x-1}}\right)$$ $$\left(\lim_{x\to1^+}\frac{1}{\sqrt{x}}\right) - \left(\lim_{x\to1^+}\frac{1}{\sqrt{x-1}}\right)$$ $$=1-\infty$$ $$=-\infty$$
(Pardon my abuse of notation.)
You can verify this graphically.
One more thing, I wanted to have a look at your attempt to combine fractions. This does not affect the answer above.
$$\frac{1}{\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}}-\frac{1}{\sqrt{x-1}}\cdot\frac{\sqrt{x-1}}{\sqrt{x-1}}$$ $$\frac{\sqrt{x}}{x}-\frac{\sqrt{x-1}}{x-1}$$ $$\frac{\sqrt{x}}{x}\cdot\frac{x-1}{x-1}-\frac{\sqrt{x-1}}{x-1}\cdot\frac{x}{x}$$ $$\frac{(x-1)\sqrt{x}-x\sqrt{x-1}}{x(x-1)}$$
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