Question

$\underset{x\to 0}{lim}\frac{{(\sin x-\tan x)}^2+{(1-\cos 2x)}^4+x^5}{7{\tan }^{7}x+{\sin }^{6}x+2{\sin }^{5}x}$ is equal to

• A. 0
• B. 1
• C. $\frac{1}{2}$
• D. 2

Answer 1

Answer: (C)

$\frac{1}{2}$

$\underset{x\to 0}{lim}\frac{{(\sin x-\tan x)}^2+{(1-\cos 2x)}^4+x^5}{7{\tan }^{7}x+{\sin }^{6}x+2{\sin }^{5}x}$
As $x\to 0\sin x,\tan x$ can be replaced with $x$
$\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1and\underset{x\to 0}{lim}\frac{\tan x}{x}=1$

$\Rightarrow \underset{x\to 0}{lim}\frac{{(\sin x-\tan x)}^2+{\left(2{\sin }^{2}x\right)}^4+x^5}{7{\tan }^{7}x+{\sin }^{6}x+2{\sin }^{5}x}$

$\Rightarrow \underset{x\to 0}{lim}\frac{{(x-x)}^2+{\left(2x^2\right)}^4+x^5}{7x^7+x^6+2x^5}$

$\Rightarrow \underset{x\to 0}{lim}\frac{16x^3+1}{7x^2+x+2}=\frac{1}{2}$