Question

Arrange the following limits in the ascending order.
a) $\underset{x\to 0}{lim}\frac{{\tan }^{4}x-{\sin }^{4}x}{x^6}$
b) $\underset{x\to 0}{lim}\frac{{\tan }^{8}x-{\sin }^{8}x}{x^5\tan x^5}$
c) $\underset{x\to 0}{lim}\frac{{\tan }^{3}x-{\sin }^{3}x}{x{\sin }^{4}x}$
d) $\underset{x\to 0}{lim}\frac{{\tan }^{5}x-{\sin }^{5}x}{x^2.{\sin }^{3}x{\tan }^{2}x}$

• A. a,b,a,d
• B. c,a,d,b
• C. a,b,d,c
• D. b,a,c,d

Answer 1

Answer: (B)

c,a,d,b

Consider, $\underset{x\to 0}{lim}\frac{(tanx-sinx)}{x^3}$
$\underset{x\to 0}{lim}\frac{sinx(1-cosx)}{x^{3}cosx}=\underset{x\to 0}{lim}\frac{\frac{sinx}{x}\cdot \frac{(1-cosx)}{x^2}}{cosx}$
$\because {lim}_{x\to 0}\frac{sinx}{x}=1\&{lim}_{x\to 0}\frac{1-cosx}{x^2}=\frac{1}{2}$
$\therefore \underset{x\to 0}{lim}\frac{(tanx-sinx)}{x^3}=\frac{1}{2}$
Using the above limit, we can evaluate the options:
$\left(a\right)\underset{x\to 0}{lim}\frac{(ta{n}^{2}x-si{n}^{2}x)(ta{n}^{2}x+si{n}^{2}x)}{x^6}$
$=\underset{x\to 0}{lim}\frac{(tanx-sinx)}{x^3}\frac{(tanx+sinx)}{x}\frac{(si{n}^2+ta{n}^{2}x)}{x^2}$
$=\frac{1}{2}\times \left(2\right)\left(2\right)=2$
$\left(b\right)\underset{x\to 0}{lim}\frac{(ta{n}^{4}x+si{n}^{4}x)(si{n}^{2}x+ta{n}^{2}x)(tanx+sinx)(tanx-sinx)}{x^4\times x^2\times x\times x^3\times \frac{tan{x}^5}{x^5}}$
$=\underset{x\to 0}{lim}\frac{(ta{n}^{4}x+si{n}^{4}x)}{x^4}\frac{(si{n}^{2}x+ta{n}^{2}x)}{x^2}\frac{(tanx+sinx)}{x}\frac{(tanx-sinx)}{x^3}$
$=2\times 2\times 2\times \frac{1}{2}=4$
$\left(c\right)\underset{x\to 0}{lim}\frac{(tanx-sinx)(ta{n}^{2}x+si{n}^{2}x+sinxtanx)}{x^5\frac{si{n}^{4}x}{x^4}}$
$=\underset{x\to 0}{lim}\frac{(tanx-sinx)}{x^3}\frac{(ta{n}^{2}x+si{n}^{2}x+sinxtanx)}{x^2}$
$=\frac{1}{2}\times 3=\frac{3}{2}$
$\left(d\right)\underset{x\to 0}{lim}\frac{(tanx-sinx)}{x^3}\frac{(ta{n}^{4}x+ta{n}^{3}xsinx+ta{n}^{2}xsi{n}^2+si{n}^{3}xtanx+si{n}^{4}x)}{x^4\frac{si{n}^{3}x}{x^3}\frac{ta{n}^{2}x}{x^2}}$
$=\frac{1}{2}\times 5=\frac{5}{2}$