Let $L:=\lim_{x\to\infty}xf(x).$ So there exists $N > 1$ such that $|xf(x)-L| \le 1$ for $x \ge N.$ Now write $$\int_1^{\infty} \frac{f(x)}{x} \ dx = \int_1^N \frac{f(x)}{x} \ dx + \int_N^{\infty} \frac{f(x)}{x} \ dx. \ \ \ \ \ \ \ \ (1)$$The first integral on the right-hand side of $(1)$ is a proper integral and so convergent. For the second integral, write $$\frac{|f(x)|}{x}=\frac{|xf(x)|}{x^2} \le \frac{|xf(x)-L|}{x^2}+\frac{|L|}{x^2} \le \frac{1+|L|}{x^2}. \ \ \ \ \ \ \ \ \ \ (2)$$So, since $\int_N^{\infty} \frac{dx}{x^2}$ is convergent, $\int_N^{\infty} \frac{f(x)}{x} \ dx$ is (absolutely) convergent too, by the comparison test. Next, put $x^t=y.$ Then $$t\int_1^a f(x^t) \ dx = \int_1^{a^t}\frac{f(y)}{y}y^{1/t}dy=\int_1^{a^t} \frac{f(y)}{y}(y^{1/t}-1) \ dy + \int_1^{a^t} \frac{f(y)}{y} \ dy$$and so $$\lim_{t\to\infty} t \int_1^a f(x^t) \ dx = \lim_{t\to\infty} \int_1^{a^t} \frac{f(y)}{y}(y^{1/t}-1) \ dy + \int_1^{\infty} \frac{f(y)}{y} \ dy.$$So we are done if we show that $$\lim_{t\to\infty} \int_1^{a^t} \frac{f(y)}{y}(y^{1/t}-1) \ dy=0. \ \ \ \ \ \ \ \ \ (3)$$To prove $(3),$ we use $(2)$ to write $$0 \le \int_1^{a^t} \frac{|f(y)|}{y}(y^{1/t}-1) \ dy \le \int_1^N \frac{|f(y)|}{y}(y^{1/t}-1) \ dy + (1+|L|)\int_N^{a^t}\frac{y^{1/t}-1}{y^2} \ dy.$$So if $M:=\max_{y \in [1,N]} |f(y)|,$ then $$0 \le \int_1^{a^t} \frac{|f(y)|}{y}(y^{1/t}-1) \ dy \le M\int_1^N \frac{y^{1/t}-1}{y} \ dy + (1+|L|)\int_N^{a^t}\frac{y^{1/t}-1}{y^2} \ dy.$$An easy calculation shows that $$\lim_{t\to\infty}\int_1^N \frac{y^{1/t}-1}{y} \ dy=\lim_{t\to\infty} \int_N^{a^t}\frac{y^{1/t}-1}{y^2} \ dy=0$$and that completes the proof of $(3)$ and the first part of your problem. For the second part of your problem, apply the first part to the function $f(x)=\frac{1}{1+x},$ which satisfies all the conditions required, to get $$ \lim_{t\to \infty} t\int_1^a \frac{dx}{1+x^t}=\int_1^{\infty} \frac{dx}{x(1+x)}=\ln 2.$$