Question

2 mark for review
$limlimits_{x\to0}\frac{6e^{4x}-2e^{3x}-4}{sin(2x)}=$
a 2
b 4
c 9
d 18

Explanation:

Step1: Check the form of the limit

When \(x
ightarrow0\), \(\lim_{x
ightarrow0}(6e^{4x}-2e^{3x}-4)=6 - 2-4=0\) and \(\lim_{x
ightarrow0}\sin(2x)=0\). So, it is in the \(\frac{0}{0}\) form. Apply L - H rule.
Differentiate the numerator and denominator.
The derivative of \(y = 6e^{4x}-2e^{3x}-4\) is \(y^\prime=24e^{4x}-6e^{3x}\) (using the formula \((e^{ax})^\prime = ae^{ax}\)).
The derivative of \(y=\sin(2x)\) is \(y^\prime = 2\cos(2x)\) (using the formula \((\sin(ax))^\prime=a\cos(ax)\)).
So, \(\lim_{x
ightarrow0}\frac{6e^{4x}-2e^{3x}-4}{\sin(2x)}=\lim_{x
ightarrow0}\frac{24e^{4x}-6e^{3x}}{2\cos(2x)}\)

Step2: Evaluate the new limit

Substitute \(x = 0\) into \(\frac{24e^{4x}-6e^{3x}}{2\cos(2x)}\)
When \(x = 0\), \(e^{4x}=e^{0}=1\), \(e^{3x}=e^{0}=1\) and \(\cos(2x)=\cos(0) = 1\)
\(\frac{24e^{4x}-6e^{3x}}{2\cos(2x)}\big|_{x = 0}=\frac{24\times1-6\times1}{2\times1}=\frac{24 - 6}{2}=\frac{18}{2}=9\)

Answer:

C. 9