use lhopitals rule to find the limit. lim x→0 x12^x / 12^x - 1 lim x→0 x12^x / 12^x - 1 = (type an exact…

Explanation:

Step1: Check indeterminate form

When (x\rightarrow0), (\lim_{x\rightarrow0}\frac{x12^{x}}{12^{x}-1}) is in the (\frac{0}{0}) form since when (x = 0), (x12^{x}=0\times12^{0}=0) and (12^{x}-1=12^{0}-1=0).

Step2: Apply L'Hopital's Rule

Differentiate the numerator and denominator. The derivative of (y = x12^{x}) using the product - rule ((uv)^\prime=u^\prime v + uv^\prime) where (u = x), (u^\prime=1), (v = 12^{x}), (v^\prime=12^{x}\ln(12)) is (y^\prime=12^{x}+x12^{x}\ln(12)). The derivative of (y = 12^{x}-1) is (y^\prime=12^{x}\ln(12)). So, (\lim_{x\rightarrow0}\frac{x12^{x}}{12^{x}-1}=\lim_{x\rightarrow0}\frac{12^{x}+x12^{x}\ln(12)}{12^{x}\ln(12)}).

Step3: Evaluate the new limit

Substitute (x = 0) into (\frac{12^{x}+x12^{x}\ln(12)}{12^{x}\ln(12)}). When (x = 0), (12^{0}+0\times12^{0}\ln(12)=1) and (12^{0}\ln(12)=\ln(12)). So (\lim_{x\rightarrow0}\frac{12^{x}+x12^{x}\ln(12)}{12^{x}\ln(12)}=\frac{1 + 0}{ \ln(12)}=\frac{1}{\ln(12)}).

Answer:

(\frac{1}{\ln(12)})