where [ ] denotes greatest integer function
Rearranging the above expression and using L'Hospital Rule in $\lim _{x \rightarrow \infty}\left(\begin{array}{c}\sin \frac{1}{\sqrt{x}} \\ \frac{-1}{x}\end{array}\right) \Rightarrow \lim _{x \rightarrow \infty}\left(\frac{\cos \frac{1}{\sqrt{x}}\left(\frac{-1}{2}\right) x^{\frac{-3}{2}}}{\left(\frac{-1}{4} x^{\frac{-5}{4}}\right)}=0\right.$
b) $\lim _{x \rightarrow \frac{\pi}{2}}(1-\sin x) \tan x(0 \times \infty$ form $)$
Rearranging and applying L'Hospital Rule $\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\cot x}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{-\cos x}{-\csc ^{2} x}=\lim _{x \rightarrow \frac{\pi}{2}} \cos x \sin ^{2} x=0$
c) $\lim _{x \rightarrow 3^{+}} \frac{[x]^{2}-9}{x^{2}-9}$
Applying L'Hospital Rule $\left(\frac{0}{0}\right)$ form
The derivative of $[x]$ at $x \rightarrow 3^{+}$ will be 0
Thus, $\lim _{x \rightarrow 3^{+}} \frac{0}{2 x}=0$
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