find the limit. use lhospitals rule if appropriate. if there is a more elementary method, consider using it…

Explanation:

Step1: Check the form at $x = 1$

Substitute $x = 1$ into $\frac{1 - x+\ln x}{1+\cos(5\pi x)}$. We get $\frac{1 - 1+\ln(1)}{1+\cos(5\pi)}=\frac{0 + 0}{1-1}=\frac{0}{0}$, so L'Hopital's Rule can be applied.

Step2: Differentiate the numerator and denominator

The derivative of the numerator $y_1=1 - x+\ln x$ is $y_1^\prime=-1+\frac{1}{x}$ using the power - rule ($\frac{d}{dx}(x^n)=nx^{n - 1}$) and $\frac{d}{dx}(\ln x)=\frac{1}{x}$. The derivative of the denominator $y_2 = 1+\cos(5\pi x)$ is $y_2^\prime=-5\pi\sin(5\pi x)$ using the chain - rule $\frac{d}{dx}(\cos(u))=-\sin(u)\cdot u^\prime$ where $u = 5\pi x$ and $u^\prime=5\pi$.

Step3: Find the new limit

We now find $\lim_{x\rightarrow1}\frac{-1+\frac{1}{x}}{-5\pi\sin(5\pi x)}$. Substitute $x = 1$ into $\frac{-1+\frac{1}{x}}{-5\pi\sin(5\pi x)}$, we get $\frac{-1 + 1}{-5\pi\sin(5\pi)}=\frac{0}{0}$, so we apply L'Hopital's Rule again.

Step4: Differentiate the new numerator and denominator

The derivative of the new numerator $y_3=-1+\frac{1}{x}= - 1+x^{-1}$ is $y_3^\prime=-x^{-2}=-\frac{1}{x^{2}}$. The derivative of the new denominator $y_4=-5\pi\sin(5\pi x)$ is $y_4^\prime=-25\pi^{2}\cos(5\pi x)$ using the chain - rule.

Step5: Find the final limit

Now find $\lim_{x\rightarrow1}\frac{-\frac{1}{x^{2}}}{-25\pi^{2}\cos(5\pi x)}$. Substitute $x = 1$ into the expression: $\frac{-\frac{1}{1^{2}}}{-25\pi^{2}\cos(5\pi)}=\frac{-1}{-25\pi^{2}\times(-1)}=-\frac{1}{25\pi^{2}}$.

Answer:

$-\frac{1}{25\pi^{2}}$