Question

question evaluate the limit: $limlimits_{x \to 1} \frac{4x - 7}{\frac{5}{3} - \frac{3x - 1}{x + 5}}$ answer attempt 1 out of 2 input box submit answer dne

Explanation:

Step1: Simplify the denominator

First, we simplify the denominator \(\frac{5}{3}-\frac{3x - 1}{x + 5}\). Find a common denominator, which is \(3(x + 5)\).
\[

$$\begin{align*} \frac{5}{3}-\frac{3x - 1}{x + 5}&=\frac{5(x + 5)-3(3x - 1)}{3(x + 5)}\\ &=\frac{5x+25 - 9x + 3}{3(x + 5)}\\ &=\frac{-4x + 28}{3(x + 5)}\\ &=\frac{-4(x - 7)}{3(x + 5)} \end{align*}$$

\]

Step2: Rewrite the original limit

Now the original limit \(\lim_{x
ightarrow1}\frac{4x - 7}{\frac{5}{3}-\frac{3x - 1}{x + 5}}\) becomes \(\lim_{x
ightarrow1}\frac{4x - 7}{\frac{-4(x - 7)}{3(x + 5)}}\). When dividing by a fraction, we multiply by its reciprocal:
\[
\lim_{x
ightarrow1}\frac{(4x - 7)\times3(x + 5)}{-4(x - 7)}
\]

Step3: Substitute \(x = 1\)

Now we substitute \(x = 1\) into the expression \(\frac{3(4x - 7)(x + 5)}{-4(x - 7)}\):
\[

$$\begin{align*} \frac{3(4\times1 - 7)(1 + 5)}{-4(1 - 7)}&=\frac{3(4 - 7)\times6}{-4\times(-6)}\\ &=\frac{3\times(-3)\times6}{24}\\ &=\frac{-54}{24}\\ &=-\frac{9}{4} \end{align*}$$

\]

Answer:

\(-\frac{9}{4}\)