Limit Calculation

**Telo** · December 25th 2012, 10:46 AM

please help me with this one to

**veileen** · December 25th 2012, 11:27 AM

If $x\rightarrow 2$ , then $5x-10\rightarrow0$ and $\sqrt[3]{x-1}-1\rightarrow0$ . So we have an indeterminate form, $\frac{0}{0}$ , use l'Hôpital's rule.

The answer is 15.

**emakarov** · December 25th 2012, 11:32 AM

A standard way to find limits of the form $\lim_{x\to a}\frac{f(x)-c^2}{\sqrt{f(x)}-c}$ where $\lim_{x\to a}f(x)=c^2$ is to multiply the numerator and the denominator by $\sqrt{f(x)}+c$ to get

$\frac{(f(x)-c^2)(\sqrt{f(x)}+c)}{(\sqrt{f(x)}-c)(\sqrt{f(x)}+c)} = \frac{(f(x)-c^2)(\sqrt{f(x)}+c)}{f(x)-c^2} = \sqrt{f(x)}+c$

where the latter expression does not have a singularity at $a$ .

For this problem, use the same idea and the fact that

$a^3-b^3=(a-b)(a^2+ab+b^2)$ (*)

Equivalently, factor out 5 and represent x - 2 in the numerator as (x - 1) - 1 and then factor it according to (*).

**Telo** · December 25th 2012, 12:12 PM

Thank you very much now i get it

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