Question

Find $\underset{x\to \infty }{lim}\frac{ax+b}{bx+a}$.

Answer 1

$L=\underset{x\to \infty }{lt}\frac{ax+b}{bx+a}$

This is a $\infty /\infty$ form

$\therefore$ Taking highest power of 'x' common and cancelling it out.

$\therefore L=\underset{x\to \infty }{lt}\frac{x}{x}\frac{(a+b/x)}{(b+a/x)}$

Now, $1/x$ becomes zero as $x\to \infty$

$\therefore$ Taking limits,

$L=\frac{a+b/\infty }{b+a/\infty }$

$\therefore L=\frac{a}{b}$.