Question

question 8 · 1 point evaluate the following limit using lhospitals rule. $lim_{x
ightarrowinfty}(1 - \frac{3}{4x})^{7x}$ provide your answer below: $lim_{x
ightarrowinfty}(1 - \frac{3}{4x})^{7x}=square$

Explanation:

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

We want to find \(\lim_{x
ightarrow\infty}\ln y=\lim_{x
ightarrow\infty}7x\ln(1 - \frac{3}{4x})\). This is in the \(0\times\infty\) form. We rewrite it as \(\lim_{x
ightarrow\infty}\frac{\ln(1-\frac{3}{4x})}{\frac{1}{7x}}\), 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 \(\lim_{x
ightarrow\infty}\frac{\ln(1-\frac{3}{4x})}{\frac{1}{7x}}=\lim_{x
ightarrow\infty}\frac{\frac{3}{4x^{2}-3x}}{-\frac{1}{7x^{2}}}\).

Step3: Simplify the limit

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

Step4: Find the original limit

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

Answer:

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