question 8 · 1 point evaluate the following limit using lhospitals rule. $lim_{x\rightarrowinfty}(1…

Explanation:

Step1: Let (y=(1 - \frac{3}{4x})^{7x}), then (\ln y = 7x\ln(1-\frac{3}{4x}))

We want to find (\lim_{x\rightarrow\infty}\ln y=\lim_{x\rightarrow\infty}7x\ln(1 - \frac{3}{4x})). This is in the (0\times\infty) form. We rewrite it as (\lim_{x\rightarrow\infty}\frac{\ln(1-\frac{3}{4x})}{\frac{1}{7x}}), which is in the (\frac{0}{0}) form so we can apply L'Hopital's rule.

Step2: Differentiate the numerator and denominator

The derivative of the numerator (u = \ln(1-\frac{3}{4x})) using the chain - rule: (u^\prime=\frac{1}{1-\frac{3}{4x}}\times\frac{3}{4x^{2}}=\frac{3}{4x^{2}-3x}). The derivative of the denominator (v=\frac{1}{7x}), (v^\prime=-\frac{1}{7x^{2}}). So (\lim_{x\rightarrow\infty}\frac{\ln(1-\frac{3}{4x})}{\frac{1}{7x}}=\lim_{x\rightarrow\infty}\frac{\frac{3}{4x^{2}-3x}}{-\frac{1}{7x^{2}}}).

Step3: Simplify the limit

(\lim_{x\rightarrow\infty}\frac{\frac{3}{4x^{2}-3x}}{-\frac{1}{7x^{2}}}=\lim_{x\rightarrow\infty}\frac{3\times7x^{2}}{-(4x^{2}-3x)}=\lim_{x\rightarrow\infty}\frac{21x^{2}}{-4x^{2}+3x}). Dividing both the numerator and denominator by (x^{2}), we get (\lim_{x\rightarrow\infty}\frac{21}{-4 + \frac{3}{x}}=-\frac{21}{4}).

Step4: Find the original limit

Since (\lim_{x\rightarrow\infty}\ln y=-\frac{21}{4}), then (y = e^{-\frac{21}{4}}) because (y = e^{\ln y}) and (\lim_{x\rightarrow\infty}y=\lim_{x\rightarrow\infty}e^{\ln y}=e^{\lim_{x\rightarrow\infty}\ln y}).

Answer:

(e^{-\frac{21}{4}})