Attachment 26349
please help me with this one to
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Attachment 26349
please help me with this one to
If $\displaystyle x\rightarrow 2$, then $\displaystyle 5x-10\rightarrow0$ and $\displaystyle \sqrt[3]{x-1}-1\rightarrow0$. So we have an indeterminate form, $\displaystyle \frac{0}{0}$, use l'Hôpital's rule.
The answer is 15.
A standard way to find limits of the form $\displaystyle \lim_{x\to a}\frac{f(x)-c^2}{\sqrt{f(x)}-c}$ where $\displaystyle \lim_{x\to a}f(x)=c^2$ is to multiply the numerator and the denominator by $\displaystyle \sqrt{f(x)}+c$ to get
$\displaystyle \frac{(f(x)-c^2)(\sqrt{f(x)}+c)}{(\sqrt{f(x)}-c)(\sqrt{f(x)}+c)} = \frac{(f(x)-c^2)(\sqrt{f(x)}+c)}{f(x)-c^2} = \sqrt{f(x)}+c$
where the latter expression does not have a singularity at $\displaystyle a$.
For this problem, use the same idea and the fact that
$\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)$ (*)
Equivalently, factor out 5 and represent x - 2 in the numerator as (x - 1) - 1 and then factor it according to (*).
Thank you very much now i get it :)