Use L’Hôpital’s rule to determine the value of

$$\underset{x\to 0}{\text{lim}}\left(\frac{{{\text{e}}^{-3x}}^{{}^2}+3\text{cos}\left(2x\right)-4}{3x^2}\right)$$

Hence find $\underset{x\to 0}{\text{lim}}\left(\frac{{\int }_0^x\left({{\text{e}}^{-3t}}^{{}^2}+3\text{cos}\left(2t\right)-4\right)\text{d}t}{{\int }_0^{x}3t^2\text{d}t}\right)$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\underset{x\to 0}{\text{lim}}\frac{{{\text{e}}^{-3x}}^{{}^2}+3\text{cos}2x-4}{3x^2}=\left(\begin{array}{c}0\\ 0\end{array}\right)$

$=\underset{x\to 0}{\text{lim}}\frac{-\text{6}x{{\text{e}}^{-3x}}^{{}^2}-6\text{sin}2x}{6x}=\left(\begin{array}{c}0\\ 0\end{array}\right)$       M1A1A1

$=\underset{x\to 0}{\text{lim}}\frac{-\text{6}{{\text{e}}^{-3x}}^{{}^2}+36x^2{{\text{e}}^{-3x}}^{{}^2}-12\text{cos}2x}{6}$      A1

= −3       A1

[5 marks]

$\underset{x\to 0}{\text{lim}}\left(\frac{{\int }_0^x\left({{\text{e}}^{-3t}}^{{}^2}+3\text{cos}2t-4\right)\text{d}t}{{\int }_0^{x}3t^2\text{d}t}\right)$ is of the form $\frac{0}{0}$

applying l’Hôpital´s rule        (M1)

$\underset{x\to 0}{\text{lim}}\frac{{{\text{e}}^{-3x}}^{{}^2}+3\text{cos}2x-4}{3x^2}$       (A1)

= −3        A1 

[3 marks]