Question

question 6 - 1 point
evaluate the following limit using lhospitals rule.
enter an exact answer.
provide your answer below:
$lim_{x
ightarrow - 2}\frac{\frac{x^{2}+7x + 10}{10}}{ln(4x + 9)}$

Explanation:

Step1: Check the form of the limit

When \(x\to - 2\), the numerator \(y_1=\frac{x^{2}+7x + 10}{10}=\frac{(x + 2)(x+5)}{10}\to0\) and the denominator \(y_2=\ln(4x + 9)\to\ln(4\times(-2)+9)=\ln(1)=0\). So, it is in the \(\frac{0}{0}\) form, and L'Hopital's rule can be applied.

Step2: Differentiate the numerator and denominator

The derivative of the numerator \(y_1=\frac{x^{2}+7x + 10}{10}\) using the power - rule \((x^n)^\prime=nx^{n - 1}\) is \(y_1^\prime=\frac{2x + 7}{10}\). The derivative of the denominator \(y_2=\ln(4x + 9)\) using the chain - rule \((\ln(u))^\prime=\frac{u^\prime}{u}\) (where \(u = 4x+9\) and \(u^\prime=4\)) is \(y_2^\prime=\frac{4}{4x + 9}\).

Step3: Calculate the new limit

The new limit is \(\lim_{x\to - 2}\frac{\frac{2x + 7}{10}}{\frac{4}{4x + 9}}=\lim_{x\to - 2}\frac{(2x + 7)(4x + 9)}{40}\).

Step4: Substitute \(x=-2\)

Substitute \(x =-2\) into \(\frac{(2x + 7)(4x + 9)}{40}\), we get \(\frac{(2\times(-2)+7)(4\times(-2)+9)}{40}=\frac{(-4 + 7)(-8 + 9)}{40}=\frac{3\times1}{40}=\frac{3}{40}\).

Answer:

\(\frac{3}{40}\)