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I really don't know how to evaluate the limit of the following function. That's the only exercise with logarithms I had so I don`t really have enough experience to know how to evaluate the limit.

$$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)$$

Use equivalents: we know that, near $0$, $ln(1+u)sim u$, so
$$lnbiggl(frac{x+7}{x+2}biggr)=lnbiggl(1+frac{5}{x+2}biggr)sim_{xto +infty}frac{5}{x+2},$$
therefore
$$xlnbiggl(frac{x+7}{x+2}biggr)sim_{xto +infty}frac{5x}{x+2}to 5.$$

Since $$log(1+t) sim t$$ as $t to 0$, we have that

$$lim_{x to infty} x log left(1 + frac 5{x+2}right) = lim_{x to infty} xcdot frac 5{x+2} = 5$$

HINT

$$xlogleft(frac{x+7}{x+2}right)=log left(1 + frac 5{x+2}right)^x$$

As an alternative way, you can also note that

$$xlnbigg(frac{x+7}{x+2}bigg) = lnbigg(frac{x+7}{x+2}bigg)^x = lnbigg(1+frac{5}{x+2}bigg)^x$$

$$= lnBiggl[bigg(1+frac{5}{x+2}bigg)^{x+2}Biggl]^{frac{x}{x+2}}$$

and

$$lim_{nto infty} bigg(1+frac{x}{n}bigg)^n = e^x$$

First note that, $$lim_{xto 0}frac{log(1+mx)}{x}=m$$ for any real $m$.

Putting, $x=1/y$ we get, $xtoinftyimplies yto 0$ and the limit becomes,
$$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)=lim_{yto0}frac{1}{y}logleft(frac{1+7y}{1+2y}right)=lim_{yto0}frac{1}{y}{log(1+7y)-log(1+2y)}=7-2=5$$

$lim_{x rightarrow infty} log(dfrac{(1+7/x)^x}{(1+2/x)^x})=$

$log dfrac{lim_{x rightarrow infty}(1+7/x)^x}{lim_{x rightarrow infty}(1+2/x)^x}=$

$log left(dfrac{e^7}{e^2}right)= 5.$

Used :

$lim_{x rightarrow infty} (1+a/x)^x=e^a$, $a$ real.

$displaystyle lim_{x to infty} x logbiggr(frac{x+7}{x+2}biggl)$ is of the indeterminate form $0 cdot infty$. We can rewrite the expression to be of the form $dfrac{infty}{frac{1}{0}}$ to be able to apply L'Hopital's Rule.

$$x log biggr(frac{x+7}{x+2}biggl)= dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} implies displaystyle lim_{x to infty} dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} = displaystyle lim_{x to infty} dfrac{frac{x+2}{x+7}}{frac{-1}{x^2}}cdot frac{d}{dx}biggr(frac{x+7}{x+2}biggl)$$

Thus the limit translates to the new equivalent simplified expression i.e. $displaystyle lim_{x to infty} dfrac{5x^2}{(x+7)(x+2)}$.

Therefore, $displaystyle lim_{x to infty} x log biggr(frac{x+7}{x+2}biggl)=5$. Cheers!