Question

use lhôpitals rule to find the limit.
lim_{x
ightarrow7}\frac{6x - 42}{6x^{2}-294}
lim_{x
ightarrow7}\frac{6x - 42}{6x^{2}-294}=square

Explanation:

Step1: Check the form

When \(x
ightarrow7\), the numerator \(6x - 42=6\times7 - 42 = 0\) and the denominator \(6x^{2}-294=6\times7^{2}-294=6\times49 - 294=294 - 294 = 0\). So, it is in the \(\frac{0}{0}\) - form, and l'Hopital's Rule can be applied.

Step2: Differentiate numerator and denominator

The derivative of the numerator \(y_1 = 6x-42\) with respect to \(x\) is \(y_1^\prime=\frac{d}{dx}(6x - 42)=6\).
The derivative of the denominator \(y_2 = 6x^{2}-294\) with respect to \(x\) is \(y_2^\prime=\frac{d}{dx}(6x^{2}-294)=12x\).

Step3: Find the new limit

We now find \(\lim_{x
ightarrow7}\frac{y_1^\prime}{y_2^\prime}=\lim_{x
ightarrow7}\frac{6}{12x}\).
Substitute \(x = 7\) into \(\frac{6}{12x}\), we get \(\frac{6}{12\times7}=\frac{6}{84}=\frac{1}{14}\).

Answer:

\(\frac{1}{14}\)