Note that $$ \begin{align} (x+a)\log\left(\frac{x+a}{x+b}\right) &=(x+a)\int_{x+b}^{x+a}\frac1t\,\mathrm{d}t\tag{1a}\\ &=(x+a)\int_{1+b/x}^{1+a/x}\frac1t\,\mathrm{d}t\tag{1b} \end{align} $$ Explanation:
$\text{(1a)}$: $\log(t)$ is the integral of $1/t$
$\text{(1b)}$: substitute $t\mapsto xt$

Next, we have the bounds $$ \overbrace{\ \ \frac{a-b}x\ \ }^{\substack{\text{width of}\\\text{domain}}}\overbrace{\frac1{1+a/x}}^{\substack{\text{minimum of}\\\text{integrand}}}\le\int_{1+b/x}^{1+a/x}\frac1t\,\mathrm{d}t\le\overbrace{\ \ \frac{a-b}x\ \ }^{\substack{\text{width of}\\\text{domain}}}\overbrace{\frac1{1+b/x}}^{\substack{\text{maximum of}\\\text{integrand}}}\tag2 $$ Multiplying $(2)$ by $x+a$ and then applying $(1)$, we get $$ a-b\le(x+a)\log\left(\frac{x+a}{x+b}\right)\le\frac{x+a}{x+b}(a-b)\tag3 $$ The Squeeze Theorem then gives $$ \lim_{x\to\infty}(x+a)\log\left(\frac{x+a}{x+b}\right)=a-b\tag4 $$

Note that $$ \begin{align} (x+a)\log\left(\frac{x+a}{x+b}\right) &=(x+a)\int_{x+b}^{x+a}\frac1t\,\mathrm{d}t\tag{1a}\\ &=(x+a)\int_{1+b/x}^{1+a/x}\frac1t\,\mathrm{d}t\tag{1b} \end{align} $$ Explanation:
$\text{(1a)}$: $\log(t)$ is the integral of $1/t$
$\text{(1b)}$: substitute $t\mapsto xt$

Next, we have the bounds $$ \overbrace{\ \ \frac{a-b}x\ \ }^{\substack{\text{width of}\\\text{domain}}}\overbrace{\frac1{1+a/x}}^{\substack{\text{minimum of}\\\text{integrand}}}\le\int_{1+b/x}^{1+a/x}\frac1t\,\mathrm{d}t\le\overbrace{\ \ \frac{a-b}x\ \ }^{\substack{\text{width of}\\\text{domain}}}\overbrace{\frac1{1+b/x}}^{\substack{\text{maximum of}\\\text{integrand}}}\tag2 $$ Multiplying $(2)$ by $x+a$ and then applying $(1)$, we get $$ a-b\le(x+a)\log\left(\frac{x+a}{x+b}\right)\le(a-b)\frac{x+a}{x+b}\tag3 $$ Graphically, we can see the function squeezed between two functions whose limits are easier to compute:

The Squeeze Theorem then gives $$ \lim_{x\to\infty}(x+a)\log\left(\frac{x+a}{x+b}\right)=a-b\tag4 $$

Note that $$ \begin{align} (x+a)\log\left(\frac{x+a}{x+b}\right) &=(x+a)\int_{x+b}^{x+a}\frac1t\,\mathrm{d}t\tag{1a}\\ &=(x+a)\int_{1+b/x}^{1+a/x}\frac1t\,\mathrm{d}t\tag{1b} \end{align} $$ Explanation:
$\text{(1a)}$: $\log(t)$ is the integral of $1/t$
$\text{(1b)}$: substitute $t\mapsto xt$

Next, we have the bounds $$ \overbrace{\ \ \frac{a-b}x\ \ }^{\substack{\text{width of}\\\text{domain}}}\overbrace{\frac1{1+a/x}}^{\substack{\text{minimum of}\\\text{integrand}}}\le\int_{1+b/x}^{1+a/x}\frac1t\,\mathrm{d}t\le\overbrace{\ \ \frac{a-b}x\ \ }^{\substack{\text{width of}\\\text{domain}}}\overbrace{\frac1{1+b/x}}^{\substack{\text{maximum of}\\\text{integrand}}}\tag2 $$ Multiplying $(2)$ by $x+a$ and then applying $(1)$, we get $$ a-b\le(x+a)\log\left(\frac{x+a}{x+b}\right)\le\frac{x+a}{x+b}(a-b)\tag3 $$ The Squeeze Theorem then gives $$ \lim_{x\to\infty}(x+a)\log\left(\frac{x+a}{x+b}\right)=a-b\tag4 $$

Note that $$ \begin{align} (x+a)\log\left(\frac{x+a}{x+b}\right) &=(x+a)\int_{x+b}^{x+a}\frac1t\,\mathrm{d}t\tag{1a}\\ &=(x+a)\int_{1+b/x}^{1+a/x}\frac1t\,\mathrm{d}t\tag{1b} \end{align} $$ Explanation:
$\text{(1a)}$: $\log(t)$ is the integral of $1/t$
$\text{(1b)}$: substitute $t\mapsto xt$

Next, we have the bounds $$ \overbrace{\ \ \frac{a-b}x\ \ }^{\substack{\text{width of}\\\text{domain}}}\overbrace{\frac1{1+a/x}}^{\substack{\text{minimum of}\\\text{integrand}}}\le\int_{1+b/x}^{1+a/x}\frac1t\,\mathrm{d}t\le\overbrace{\ \ \frac{a-b}x\ \ }^{\substack{\text{width of}\\\text{domain}}}\overbrace{\frac1{1+b/x}}^{\substack{\text{maximum of}\\\text{integrand}}}\tag2 $$ Multiplying $(2)$ by $x+a$ and then applying $(1)$, we get $$ a-b\le(x+a)\log\left(\frac{x+a}{x+b}\right)\le\frac{x+a}{x+b}(a-b)\tag3 $$ The Squeeze Theorem then gives $$ \lim_{x\to\infty}(x+a)\log\left(\frac{x+a}{x+b}\right)=a-b\tag4 $$