Existence of limit of a function from existence of another limit of a function 
Let $a \in \mathbb{R}$ and suppose that $f:\mathbb{R} \to \mathbb{R}$
  is a function such that $f(x) \ge 1,\ \forall x \in \mathbb{R}$. If
  $\lim_{x \to a}\bigg[\frac{1}{\sqrt{f(x)}} + \sqrt{f(x)}\bigg]$
  exists, then prove that $\lim_{x \to a} f(x)$ exists.
HINT: If $f(x) \ge B > 0$ and $(\lim_{x \to a} f(x))^2$ exists then $\lim_{x \to a}(f(x))^2$ exists.

Since we are given that $\lim_{x \to a}\bigg[\frac{1}{\sqrt{f(x)}} + \sqrt{f(x)}\bigg]$ exists, we can multiply this limit by itself to have
$$\lim_{x \to a}\bigg[\frac{1}{\sqrt{f(x)}} + \sqrt{f(x)}\bigg]\cdot\lim_{x \to a}\bigg[\frac{1}{\sqrt{f(x)}} + \sqrt{f(x)}\bigg] = \bigg(\lim_{x \to a}\bigg[\frac{1}{\sqrt{f(x)}} + \sqrt{f(x)}\bigg]\bigg)^2$$
exists. Therefore, by hint
$$\lim_{x \to a}\bigg[\bigg(\frac{1}{\sqrt{f(x)}} + \sqrt{f(x)}\bigg)^2\bigg] = \lim_{x \to a}\bigg[\frac{1}{f(x)} + f(x)+2\bigg] = 2+\lim_{x \to a}\bigg(\frac{1}{f(x)} + f(x)\bigg)$$
exists. We also know that $0<\frac{1}{f(x)} \le 1$ is bounded and $f(x) \ge 1$. But I am stuck here since I can't separate $\lim_{x \to a}\big(\frac{1}{f(x)} + f(x)\big)$ as $\lim_{x \to a}\frac{1}{f(x)} + \lim_{x \to a} f(x)$ as I don't know whether they separately exist or not. I thought about proof bu contradiction by assuming $\lim_{x \to a}f(x)$ does not exist but could not go any further. Any hint or advice is appreciated. Thank you in advance.

Let's write $u=f(x), v=\sqrt{f(x)} $. Then it is known that $u\geq 1,v\geq 1$ and $v+(1/v)=w$ tends to a limit $l$. Next note that $w=v+(1/v)\geq 2$ via AM-GM inequality and hence $l\geq 2$.
Solving for $v$ we get $$v=\frac{w\pm\sqrt{w^2-4}}{2}$$ Since $v\geq 1$ the $+$ sign must be choosen above. Hence we have $$v=\frac{w+\sqrt{w^2-4}}{2}$$ and thus by continuity of square/square root function $v$ tends to $$\frac{l+\sqrt{l^2-4}}{2}$$ Then by continuity of square function, $u=f(x) =v^2$ tends to the square of this number. 
