• $\lim_{x \to 0} \frac{\text{arctan}(\cos x) - \frac{\pi}{4}}{x^2} = \left(\frac{0}{0}\right)$ A1

• $= \lim_{x \to 0} \frac{-\sin x}{1+\cos^2 x} \cdot \frac{1}{2x}$ A1

Award A1 for a correct numerator and A1 for a correct denominator.

• Recognizes to apply l'Hôpital's rule again M1

• $= \lim_{x \to 0} \frac{-\sin x}{1+\cos^2 x} \cdot \frac{1}{2x} = \left(\frac{0}{0}\right)$

Award M0 if their limit is not the indeterminate form $\frac{0}{0}$.

Method #1

• $= \lim_{x \to 0} \frac{-\cos x (1+\cos^2 x) - 2 \sin^2 x\cos x}{(1+\cos^2 x)^2}$ A1A1

Award A1 for a correct first term in the numerator and A1 for a correct second term in the numerator.

Method #2

• $\lim_{x \to 0} \frac{-\cos x}{2(1+\cos^2 x)} - \frac{4x\sin x\cos x}{(1+\cos^2 x)}$ A1A1

Award A1 for a correct numerator and A1 for a correct denominator.

Then

• Substitutes $x=0$ into the correct expression to evaluate the limit A1

The final A1 is dependent on all previous marks.

• $=-\frac{1}{4}$ AG

6 marks total

• As $\lim_{x\to0}x^2 = 0$, the indeterminate form $\frac{0}{0}$ is required for the limit to exist. M1

• $\lim_{x\to0} \left(\arctan(\cos x) - m\right) = 0$

• $\arctan 1 - m = 0$ (as $m = \arctan 1$) A1

• So $m = \frac{\pi}{4}$

2 marks total