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$\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^2}$ is equal to
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$\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^2}, \quad\left(\right.$ 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
$\lim _{x \rightarrow 0} \frac{2 \sec x \cdot \sec x \cdot \tan x+\sin x}{2}$
$=\frac{2 \cdot 1 \cdot 1 \cdot 0+0}{2}=\frac{0}{2}=0$
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
$\lim _{x \rightarrow 0} \frac{2 \sec x \cdot \sec x \cdot \tan x+\sin x}{2}$
$=\frac{2 \cdot 1 \cdot 1 \cdot 0+0}{2}=\frac{0}{2}=0$
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