Is this correct way to find this limit?

Question

I have to find limit of the following

$(a^\sqrt{x}-a^\frac{1}{\sqrt{x}})/(a^\sqrt{x}+a^\frac{1}{\sqrt{x}})\;$ as $x$ tends to $0$.

My attempt: $(a^\frac{x-1}{\sqrt{x}}-1)/(a^\frac{x-1}{\sqrt{x}}+1)$

Now let $\frac{x-1}{\sqrt{x}}=t$ Then as $x$ tends to 0 $t$ tends negative infinity. So my limit is -1. Is this answer correct? Thanks in advance

Answer 1

Your approach is good but $a^{t} \to 0$ if $ a>1$, it tends to $1$ if $0<a<1$

The limit is $1$ if $0 <a<1$ and $-1$ if $a > 1$. For $a=1$ it is clearly $0$.

[We are taking limit as $ x \to 0$ through positive values of $x$ since the function is not defined for $x<0$].