Question

If $\underset{x\to 0}{lim}\frac{a+e^{1/x}}{1+b{e}^{1/x}}=2$,then $(a,b)$=

Answer 1

$\underset{x\to 0}{Lim}\frac{a+e^{1/x}}{1+b{e}^{1/x}}$

When $x\to 0;e^{1/x}\to \infty$

Thus this is $\frac{\infty }{\infty }$ indeterminate form.

Applying L'Hospitals rule,

$\underset{x\to 0}{Lim}\frac{e^{1/x}\frac{-1}{x^2}}{b{e}^{1/x}\frac{-1}{x^2}}=2$

Thus,$\frac{1}{b}=2$

$\therefore b=\frac{1}{2}$

And $aϵR$