Let f(x)=cases(x-1)^12-x, & x1, x2, & x=2cases.

Question

Let $f(x)=\begin{cases}(x-1)^{\frac12-x}, & x>1,\ x
e2\k, & x=2\end{cases}$. The value of $k$ for which $f$ is continuous at $x=2$ is

Accepted Answer

$\displaystyle\lim_{x	o2}(x-1)^{1/2-x}=e^\ell$ where $\ell=\displaystyle\lim_{x	o2}\left(\frac12-xight)\ln(x-1)$; writing $\ln(x-1)\approx(x-2)$ near $x=2$, $\ell=\displaystyle\lim_{x	o2}-\frac{x-2}{x-2}\cdot$(appropriate form) $=-1$. So $k=e^{-1}$.

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