Let f be defined on (0, infinity) by f(x) = (3 + sqrt(x)) / (4 sqrt(x)). Find the limit of f(x) and the limit of f(f(x)) as x tends to infinity.

Step-by-step solution

1. 1

   Split the fraction instead of fighting the indeterminate form. Written as it stands, $f(x)$ is an $\frac{\infty}{\infty}$ form. Dividing each term of the numerator by the denominator removes the ambiguity outright:

   $f(x)=\frac{3}{4\sqrt{x}}+\frac{\sqrt{x}}{4\sqrt{x}}=\frac{3}{4\sqrt{x}}+\frac{1}{4}$

   This rewriting is legitimate for every $x>0$, where $\sqrt{x}\neq 0$.

2. 2

   Read off the first limit. As $x\to+\infty$ we have $\sqrt{x}\to+\infty$, so $\dfrac{3}{4\sqrt{x}}\to 0$ and

   $\lim_{x\to+\infty}f(x)=0+\frac{1}{4}=\frac{1}{4}$

   The line $y=\tfrac14$ is a horizontal asymptote of the graph, approached from above since $\tfrac{3}{4\sqrt{x}}>0$ always.

3. 3

   Do not substitute $+\infty$ into the outer function. For $f(f(x))$ the inner value does not go to infinity — it goes to $\tfrac14$. The correct tool is the composition rule for limits: if $f(x)\to L$ as $x\to+\infty$, and $f$ is continuous at $L$, then

   $\lim_{x\to+\infty}f\bigl(f(x)\bigr)=f(L)$

4. 4

   Check that the rule applies. Here $L=\tfrac14$, which lies inside the domain $\left]0,+\infty\right[$, and $f$ is continuous there (it is built from $\sqrt{x}$ and a quotient whose denominator $4\sqrt{x}$ never vanishes on that interval). Moreover $f(x)>\tfrac14>0$ for every $x>0$, so the inner values genuinely stay inside the domain of the outer $f$ — the composition is well defined near infinity.

5. 5

   Evaluate the outer function at $L=\tfrac14$. Since $\sqrt{1/4}=\tfrac12$:

   $f\!\left(\frac{1}{4}\right)=\frac{3+\frac{1}{2}}{4\cdot\frac{1}{2}}=\frac{\frac{7}{2}}{2}=\frac{7}{4}$

   $\boxed{\lim_{x\to+\infty}f(x)=\frac{1}{4},\qquad \lim_{x\to+\infty}f\bigl(f(x)\bigr)=\frac{7}{4}}$

6. 6

   Confirm numerically. At $x=10^4$: $f(x)=0.2575$ and $f(f(x))=1.72799$. At $x=10^8$: $f(x)=0.250075$ and $f(f(x))=1.749775$. The values close in on $\tfrac14=0.25$ and $\tfrac74=1.75$, exactly as the continuity argument predicts.