Respuesta :
Answer:
Therefore
[tex]lim_{x \to 1}\frac{1-x+lnx}{1+ cos7\pi x}[/tex] [tex]=-\frac{1}{49\pi^2}[/tex]
Step-by-step explanation:
L Hospital's Rule:
[tex]lim_{x\to a}\frac{f(x)}{g(x)}[/tex] , when the value of [tex]\frac{f(x)}{g(x)}[/tex] at a are such as [tex]\frac{0}{0}[/tex] or [tex]\frac{\infty}{\infty}[/tex]
then we can write
[tex]lim_{x\to a}\frac{f(x)}{g(x)} =lim_{x\to a}\frac{f'(x)}{g'(x)}[/tex] where f'(x) and g'(x) are first order derivative.
Given that,
[tex]lim_{x \to 1}\frac{1-x+lnx}{1+ cos7\pi x}[/tex]
Putting x=1 in [tex]\frac{1-x+lnx}{1+ cos7\pi x}[/tex] we get [tex]\frac{1-1+ln1}{1+cos 7\pi}=\frac{0}{0}[/tex]
Therefore it satisfies the L Hospital's Rule. then differential denominator and numerator with respect to x
=[tex]lim_{x\to 1} \frac{-1+\frac{1}{x}}{-7\pi sin 7\pi x}[/tex]
Again putting x= 1 in [tex]\frac{-1+\frac{1}{x}}{-7\pi sin 7\pi x}[/tex] it gives [tex]\frac{0}{0}[/tex]. Again differential denominator and numerator with respect to x.
[tex]=lim_{x \to 1} \frac{-\frac{1}{x^2}}{-49\pi^2cos 7\pi x}[/tex]
Now putting x =1
[tex]=\frac{-\frac{1}{1^2}}{-49\pi^2cos 7 \pi}[/tex]
[tex]=\frac{-1}{-49\pi^2(-1)}[/tex]
[tex]=-\frac{1}{49\pi^2}[/tex]
Therefore
[tex]lim_{x \to 1}\frac{1-x+lnx}{1+ cos7\pi x}[/tex] [tex]=-\frac{1}{49\pi^2}[/tex]
