Find the limit of the function using l'Hopital's Rule?

Question

$$ \lim _{x\to 6}\left(\frac{x-5}{x-6}- \frac{1}{\ln(x-5)}\right)$$

Am not sure what $f(x)$ and $g(x)$ will be in this problem.

$$ \frac{(x-5)\ln(x-5)-x+6}{(x-6)\ln(x-5)}$$

Answer 1

Note that to can apply l’Hopital, we need to manipulate the given expression to the form $$\frac{f(x)}{g(x)}=\frac{(x-5)\log(x-5)-(x-6)}{(x-6)\log(x-5)}\to \frac{0}{0}$$

and the derivative of the numerator is

$$((x-5)\log(x-5)-(x-6))'=\log(x-5)+1-1=\log(x-5)$$

and for the denominator

$$((x-6)\log(x-5))'=\log(x-5)+\frac{x-6}{x-5}$$

thus

$$\frac{f'(x)}{g'(x)}=\frac{\log(x-5)}{\log(x-5)+\frac{x-6}{x-5}}\to\frac{0}{0}$$

Then we need to derive again to obtain finally

$$\frac{f''(x)}{g''(x)}=\frac{\frac{1}{x-5}}{\frac{1}{x-5}+\frac{1}{(x-5)^2}}\to\frac{1}{2}$$

Answer 2

Notice that \begin{eqnarray}\lim_{x\to 6}\left(\frac{x-5}{x-6}-\frac{1}{\ln(x-5)}\right)&=&\lim_{x\to 6}\frac{(x-5)\ln(x-5)-(x-6)}{(x-6)\ln(x-5)}\\ &=& \lim_{x\to 6}\frac{\ln(x-5)+1-1}{\ln(x-5)+\frac{x-6}{x-5}}\\ &=& \lim_{x\to 6}\frac{(x-5)^{-1}}{(x-5)^{-1}+\frac{(x-5)-(x-6)}{(x-5)^2}}\\ &=& \lim_{x\to 6}\frac{(x-5)^{-1}}{(x-5)^{-1}+\frac{1}{(x-5)^2}}\\ &=& \frac{1}{2}. \end{eqnarray}

Here we applied l'Hopital's rule in the second and third equality.

Answer 3

$$\lim_{x\rightarrow6}\left(\frac{x-5}{x-6}- \frac{1}{\ln(x-5)}\right)=\lim_{x\rightarrow6}\frac{(x-5)\ln(x-5)-x+6}{(x-6)(\ln(x-5)}=$$ $$=\lim_{x\rightarrow6}\frac{\ln(x-5)}{\ln(x-5)+\frac{x-6}{x-5}}=\lim_{x\rightarrow6}\frac{\frac{1}{x-5}}{\frac{1}{x-5}+\frac{1}{(x-5)^2}}=\frac{1}{2}$$

Answer 4

Put $t=x-6$

the function becomes

$$\frac {t+1}{t}-\frac {1}{\ln (1+t)}=$$ $$\frac {(t+1)\ln (t+1)-t}{t\ln (1+t)} $$

the derivatives are easier.