Science:Math Exam Resources/Courses/MATH104/December 2016/Question 08 (b)

MATH104 December 2016

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Question 08 (b)
Consider the following function: $f(x)={\begin{cases}4-x^{2},&x<1\\\ln x,&1<x<3\\e^{x},&x>3\end{cases}}$ Evaluate the limit $\lim _{x\to 1^{-}}f(f(x))$.

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall that $x\to 1^{-}$ means that $x$ approaches to $1$ from the left. In other words, we consider $x$ which is close to $1$ and satisfies $x<1$. Find the piece of function $f$ which lies on the region $x<1$ and determine what value $f(x)$ approaches as $x\to 1^{-}$, and from which direction (left/right) does $f(x)$ approach this value .

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Recall that $x\to 1^{-}$ means that $x$ approaches to $1$ from the left. In other words, we consider $x$ which is close to $1$ and satisfies $x<1$. For such $x$, the function $f$ has the value $f(x)=4-x^{2}$. This implies that $\lim _{x\to 1^{-}}f(f(x))=\lim _{x\to 1^{-}}f(4-x^{2})$. Then, we can observe that as $x\to 1^{-}$, the function $4-x^{2}$ approaches to $3$ from the right. i.e., $4-x^{2}\to 3^{+}$. This is because for $x<1$, we have $4-x^{2}>3$. Therefore, denoting $y=4-x^{2}$, the limit becomes $\lim _{x\to 1^{-}}f(4-x^{2})=\lim _{y\to 3+}f(y)$. Finally, on the interval $y>3$, we have $f(y)=e^{y}$, and therefore $\lim _{y\to 3+}f(y)=\lim _{y\to 3+}e^{y}=e^{3}$. Combining all the equations, we can evaluate the limit; $\lim _{x\to 1^{-}}f(f(x))=\color {blue}e^{3}$.