Q73 / 160 · TS EAMCET 2010 · Mathematics

$\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^2}$ is equal to

  • Correct answer
    A
    0
  • B
    1
  • C
    $\frac{1}{2}$
  • D
    $-\frac{1}{2}$

Solution

$\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^2}, \quad\left(\right.$ form $\left.\frac{0}{0}\right)$ By 'L' Hospital Rule $\lim _{x \rightarrow 0} \frac{\sec ^2 x-\cos x}{2 x}, \quad\left(\text { form } \frac{0}{0}\right)$ Again, by 'L' Hospital Rule $\lim _{x \rightarrow 0} \frac{2 \sec x \cdot \sec x \cdot \tan x+\sin x}{2}$ $=\frac{2 \cdot 1 \cdot 1 \cdot 0+0}{2}=\frac{0}{2}=0$

Hence, the correct answer is option: 'A'.

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