Question

Which of the following limits vanish?
where $[\cdot ]$ denotes greatest integer function.

• A. $\underset{x\to \infty }{lim}{x}^{\frac{1}{4}}sin\frac{1}{\sqrt{x}}$
• B. $\underset{x\to \frac{\pi }{2}}{lim}(1-sinx)\cdot tanx$
• C. $\underset{x\to \infty }{lim}\frac{2x^2+3}{x^2+x-5}\cdot sgn\left(x\right)$
• D. $\underset{x\to 3^{+}}{lim}\frac{{\left[x\right]}^2-9}{x^2-9}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\underset{x\to \infty }{lim}{x}^{\frac{1}{4}}sin\frac{1}{\sqrt{x}}$
B $\underset{x\to \frac{\pi }{2}}{lim}(1-sinx)\cdot tanx$
D $\underset{x\to 3^{+}}{lim}\frac{{\left[x\right]}^2-9}{x^2-9}$

a) $\underset{x\to \infty }{lim}{x}^{\frac{1}{4}}\sin \frac{1}{\sqrt{x}}$($0\times$ $\infty$ form)
Rearranging the above expression and using L'Hospital Rule in
${lim}_{x\to \infty }\left(\frac{\sin \frac{1}{\sqrt{x}}}{x^{\frac{-1}{4}}}\right)\Rightarrow \underset{x\to \infty }{lim}\left(\frac{\cos \frac{1}{\sqrt{x}}\left(\frac{-1}{2}\right)x^{\frac{-3}{2}}}{\left(\frac{-1}{4}{x}^{\frac{-5}{4}}\right)}\right)=0$
b)$\underset{x\to \frac{\pi }{2}}{lim}(1-\sin x)\tan x$($0\times \infty$ form)
Rearranging and applying L'Hospital Rule
${lim}_{x\to \frac{\pi }{2}}\frac{1-\sin x}{\cot x}=\underset{x\to \frac{\pi }{2}}{lim}\frac{-\cos x}{-{csc}^{2}x}=\underset{x\to \frac{\pi }{2}}{lim}\cos x{sin}^{2}x=0$
c)Since $sgn\left(x\right)=1$ as $x\to \infty$
Thus, $\underset{x\to \infty }{lim}\frac{2x^3+3}{x^2+x-5}$
Taking highest power of $x$ common in both numerator in denominator
$=\underset{x\to \infty }{lim}\frac{x^2(2+\frac{3}{x^2})}{x^2(1+\frac{1}{x}-\frac{5}{x^2})}=2$
d)$\underset{x\to 3^{+}}{lim}\frac{{\left[x\right]}^2-9}{x^2-9}$
Applying L'Hospital Rule $\left(\frac{0}{0}\right)$ form
The derivative of $\left[x\right]$ at $x\to 3^{+}$ will be $0$
Thus, $\underset{x\to 3^{+}}{lim}\frac{0}{2x}=0$