Question

question 12
evaluate the limit: \\(\lim\limits_{c\to -\infty} \frac{3c^4 + 4c^2 - 3}{7c^4 - c + 1}\\)
\\(\bigcirc\\) 3/7
\\(\bigcirc\\) 0
\\(\bigcirc\\) no correct answer choice is given.
\\(\bigcirc\\) \\(-\infty\\)
\\(\bigcirc\\) \\(\infty\\)

Explanation:

Step1: Divide numerator and denominator by \( c^4 \)

For the limit \( \lim_{c \to -\infty} \frac{3c^4 + 4c^2 - 3}{7c^4 - c + 1} \), we divide each term in the numerator and the denominator by the highest power of \( c \) in the denominator, which is \( c^4 \).

So the numerator becomes: \( \frac{3c^4}{c^4}+\frac{4c^2}{c^4}-\frac{3}{c^4}=3 + \frac{4}{c^2}-\frac{3}{c^4} \)

The denominator becomes: \( \frac{7c^4}{c^4}-\frac{c}{c^4}+\frac{1}{c^4}=7-\frac{1}{c^3}+\frac{1}{c^4} \)

Now the limit is \( \lim_{c \to -\infty} \frac{3 + \frac{4}{c^2}-\frac{3}{c^4}}{7-\frac{1}{c^3}+\frac{1}{c^4}} \)

Step2: Evaluate the limit as \( c \to -\infty \)

As \( c \to -\infty \), terms with \( c \) in the denominator (i.e., \( \frac{4}{c^2},\frac{3}{c^4},\frac{1}{c^3},\frac{1}{c^4} \)) will approach 0.

So we substitute the limits of these terms:

\( \lim_{c \to -\infty} 3 + \frac{4}{c^2}-\frac{3}{c^4}=3 + 0 - 0 = 3 \)

\( \lim_{c \to -\infty} 7-\frac{1}{c^3}+\frac{1}{c^4}=7 - 0 + 0 = 7 \)

Thus the limit of the fraction is \( \frac{3}{7} \)

Answer:

A. \( \frac{3}{7} \)