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 form of limit

As $x\rightarrow0$, we have $\frac{0\times12^{0}}{12^{0}-1}=\frac{0}{0}$, so L'Hopital's Rule applies.

Step2: Differentiate numerator and denominator

The derivative of $y = x12^{x}$ using the product - rule $(uv)^\prime=u^\prime v + uv^\prime$ where $u = x$ and $v = 12^{x}$. $u^\prime=1$ and $v^\prime=12^{x}\ln(12)$, so $y^\prime=12^{x}+x12^{x}\ln(12)$. The derivative of $y = 12^{x}-1$ is $y^\prime=12^{x}\ln(12)$.

Step3: Find new limit

$\lim_{x\rightarrow0}\frac{12^{x}+x12^{x}\ln(12)}{12^{x}\ln(12)}=\lim_{x\rightarrow0}\left(\frac{12^{x}}{12^{x}\ln(12)}+\frac{x12^{x}\ln(12)}{12^{x}\ln(12)}\right)$.

Step4: Simplify and evaluate limit

$\lim_{x\rightarrow0}\left(\frac{1}{\ln(12)} + x\right)=\frac{1}{\ln(12)}+0=\frac{1}{\ln(12)}$.

Answer:

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