Answer to Question #223043 in Calculus for Sandralyine

Question

Evaluate the limit as x turns to 2⁺

lim {sin[In (x-1)] / √(x-2)}

Answer 1

$\lim _{x \rightarrow 2^{+}}\left\{\frac{\sin [\ln (x-1)]}{\sqrt{x-2}}\right\}$

This is $\frac{0}{0}$ form as  $\sin [\ln (2-1)]=\sin [\ln 1]=\sin 0=0 \sqrt{2-2}=0$ and $\sqrt{2-2}=0$

So, using L' Hospital rule, 

$\lim _{x \rightarrow 2^{+}}\left\{\frac{\sin [\ln (x-1)]}{\sqrt{x-2}}\right\}\\=\lim _{x \rightarrow 2^{+}} \frac{\frac{d}{d x}(\sin [\ln (x-1)])}{\frac{d}{d x}(\sqrt{x-2})}\\=\lim _{x \rightarrow 2^{+}} \frac{\cos [\ln (x-1)] \cdot \frac{d}{d x}(\ln (x-1))}{\frac{1}{2}(x-2)^{-\frac{1}{2}}}$

$=\lim _{x \rightarrow 2^{+}} 2(x-2)^{\frac{1}{2}} \cdot \cos [\ln (x-1)] \cdot\left(\frac{1}{x-1}\right)\\=\lim _{h \rightarrow 0} 2(h+2-2)^{\frac{1}{2}} \cdot \cos [\ln (h+2-1)] \cdot\left(\frac{1}{h+2-1}\right)$

$=\lim _{h \rightarrow 0} 2(h)^{\frac{1}{2}} \cdot \cos [\ln (h+1)] \cdot\left(\frac{1}{h+1}\right)\\=2 \times 0 \times \cos [\ln (1)] \times 1\\=2 \times 0 \times \cos [0] \times 1=0$