Mathematics - Limits Question with Solution | TestHub

Consider $\underset{x\to 0}{\lim }\frac{\cos \left(\sin x\right)-\cos x}{x^4} \left(\frac{0}{0} \text{form}\right)$

$\underset{x\to 0}{\lim }\frac{2\sin \left(\frac{\sin x+x}{2}\right)·\sin \left(\frac{x-\sin x}{2}\right)}{x^4}$

$=\underset{x\to 0}{\lim }2\left[\frac{\sin \left(\frac{\sin x+x}{2}\right)}{\left(\frac{\sin x+x}{2}\right)}\right]\left[\frac{\sin \left(\frac{x-\sin x}{2}\right)}{\left(\frac{x-\sin x}{2}\right)}\right]\times \left(\frac{\sin x+x}{2}\right)\times \left(\frac{x-\sin x}{2}\right)\times \frac{1}{x^4}$

$\underset{x\to 0}{\lim }2\left[\frac{\sin \left(\frac{\sin x+x}{2}\right)}{\left(\frac{\sin x+x}{2}\right)}\right]\left[\frac{\sin \left(\frac{x-\sin x}{2}\right)}{\left(\frac{x-\sin x}{2}\right)}\right]\times \left(\frac{x^2-{\sin }^{2}x}{4x^4}\right)$

$\underset{x\to 0}{\lim }2\times \left(\frac{x^2-{\sin }^{2}x}{4x^4}\right) \left(\frac{0}{0} \text{form}\right) \left(\because \underset{t\to 0}{\lim }\frac{\sin t}{t}=1\right)$

Applying L'Hospital's Rule,

$\underset{x\to 0}{\lim }2\times \left(\frac{2x-2\sin x\cos x}{4·4x^3}\right)=\underset{x\to 0}{\lim }\left(\frac{2x-\sin 2x}{8x^3}\right) \left(\frac{0}{0} \text{form}\right)$

$\underset{x\to 0}{\lim }\left(\frac{2-2\cos 2x}{24x^2}\right) \left(\frac{0}{0} \text{form}\right)$

$\underset{x\to 0}{\lim }\left(\frac{4\sin 2x}{48x}\right)=\underset{x\to 0}{\frac{1}{6}\lim }\left(\frac{\sin 2x}{2x}\right)=\frac{1}{6}$