evaluate the following limit using lhôpitals rule. lim x→0 8 sin 6x / 7x use lhôpitals rule to rewrite the…

Explanation:

Step1: Check indeterminate form

When (x\rightarrow0), (\lim_{x\rightarrow0}\frac{8\sin6x}{7x}) is in (\frac{0}{0}) form since (\sin(0) = 0). So we can apply L'Hopital's Rule.

Step2: Differentiate numerator and denominator

The derivative of (y = 8\sin6x) using the chain - rule ((\sin u)^\prime=\cos u\cdot u^\prime) where (u = 6x) and (y^\prime=8\times6\cos6x = 48\cos6x). The derivative of (y = 7x) is (y^\prime=7). So (\lim_{x\rightarrow0}\frac{8\sin6x}{7x}=\lim_{x\rightarrow0}\frac{48\cos6x}{7}).

Step3: Evaluate the new limit

Substitute (x = 0) into (\frac{48\cos6x}{7}). Since (\cos(0)=1), we have (\lim_{x\rightarrow0}\frac{48\cos6x}{7}=\frac{48\cos(0)}{7}=\frac{48\times1}{7}=\frac{48}{7}).

Answer:

(\frac{48}{7})