question 6 - 1 point\nevaluate the following limit using lhospitals rule.\nenter an exact answer.\nprovide…

question 6 - 1 point\nevaluate the following limit using lhospitals rule.\nenter an exact answer.\nprovide your answer below:\n$lim_{x\rightarrow - 2}\frac{\frac{x^{2}+7x + 10}{10}}{ln(4x + 9)}$

question 6 - 1 point\nevaluate the following limit using lhospitals rule.\nenter an exact answer.\nprovide your answer below:\n$lim_{x\rightarrow - 2}\frac{\frac{x^{2}+7x + 10}{10}}{ln(4x + 9)}$

Answer

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})