<s>[INST] Limit of $a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}$ when $x\to\infty$, for given real numbers $a$, $b$, $c$ 
Find the limit of $a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}$ when $x\to\infty$, for given real numbers $a$, $b$, $c$.

I would like to see a solving method without l'Hopital or Taylor expansion.
 [/INST] \begin{align*}
  L &= \lim_{x\to \infty} (a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}) \\[5pt]
  &= \lim_{x\to \infty} \sqrt{x}
     \left(
       a\sqrt{1+\frac{1}{x}}+2b\sqrt{1+\frac{1}{2x}}+3c\sqrt{1+\frac{1}{3x}}
     \right) \\[5pt]
  &= \lim_{x\to \infty} \sqrt{x}
  \left[
     a\left( 1+\frac{1}{2x} \right)+
    2b\left( 1+\frac{1}{4x} \right)+
    3c\left( 1+\frac{1}{6x} \right)
  \right] \\[5pt]
  &=
  \left \{
    \begin{array}{ccl}
      \infty &,& a+2b+3c \ne 0 \\
      0 &, & a+2b+3c=0
    \end{array}
  \right.
\end{align*}
For one case of $(a,b,c)=(1,1,-1) \implies a+2b+3c=0$.
See the comparison of 
$\color{blue}{y=a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}}$ and 
$\color{red}{y=\dfrac{a+b+c}{2\sqrt{x}}}$ below:

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