Question

1. if $f(x)=\

$$\begin{cases}\\ln x & \\text{for } 0 < x \\leq 2 \\\\ x^2 \\ln 2 & \\text{for } 2 < x \\leq 4,\\end{cases}$$

$ then $\lim\limits_{x \to 2} f(x)$ is
(a) $\ln 2$ (b) $\ln 8$ (c) $\ln 16$ (d) $4$ (e) nonexistent

Explanation:

Step1: Find left - hand limit

To find the left - hand limit as \(x ightarrow2^{-}\), we use the part of the piece - wise function for \(0 < x\leq2\), which is \(f(x)=\ln x\).
The left - hand limit \(\lim_{x ightarrow2^{-}}f(x)=\lim_{x ightarrow2^{-}}\ln x\).
Since the natural logarithm function \(y = \ln x\) is continuous at \(x = 2\) (the domain of \(\ln x\) is \(x>0\) and it is continuous on its domain), we can directly substitute \(x = 2\) into \(\ln x\). So \(\lim_{x ightarrow2^{-}}f(x)=\ln2\).

Step2: Find right - hand limit

To find the right - hand limit as \(x ightarrow2^{+}\), we use the part of the piece - wise function for \(2 < x\leq4\), which is \(f(x)=x^{2}\ln2\).
The right - hand limit \(\lim_{x ightarrow2^{+}}f(x)=\lim_{x ightarrow2^{+}}x^{2}\ln2\).
Since the function \(y=x^{2}\ln2\) is a polynomial - like function (a constant multiple of a polynomial) and is continuous everywhere, we can substitute \(x = 2\) into \(x^{2}\ln2\).
Substituting \(x = 2\) into \(x^{2}\ln2\), we get \((2)^{2}\ln2=4\ln2=\ln2^{4}=\ln16\)? Wait, no, wait. Wait, let's recalculate:
Wait, \(x = 2\), so \(x^{2}\ln2=(2)^{2}\ln2 = 4\ln2\). But also, \(\ln2\) (from left - hand limit) and \(4\ln2\) (from right - hand limit) - Wait, no, I made a mistake. Wait, no:
Wait, the left - hand limit: when \(x ightarrow2^{-}\), \(f(x)=\ln x\), so \(\lim_{x ightarrow2^{-}}f(x)=\ln2\).
The right - hand limit: when \(x ightarrow2^{+}\), \(f(x)=x^{2}\ln2\), so \(\lim_{x ightarrow2^{+}}f(x)=(2)^{2}\ln2 = 4\ln2\).
Wait, but \(\ln2 eq4\ln2\), so the limit does not exist? Wait, no, wait the original problem: Wait, maybe I misread the function. Wait, the function is \(f(x)=\begin{cases}\ln x&0 < x\leq2\\x^{2}\ln2&2 < x\leq4\end{cases}\)
Wait, left - hand limit: as \(x\) approaches \(2\) from the left (\(x ightarrow2^{-}\)), \(f(x)=\ln x\), so \(\lim_{x ightarrow2^{-}}f(x)=\ln2\).
Right - hand limit: as \(x\) approaches \(2\) from the right (\(x ightarrow2^{+}\)), \(f(x)=x^{2}\ln2\), so \(\lim_{x ightarrow2^{+}}f(x)=(2)^{2}\ln2 = 4\ln2\).
Since \(\lim_{x ightarrow2^{-}}f(x)=\ln2\) and \(\lim_{x ightarrow2^{+}}f(x)=4\ln2\), and \(\ln2 eq4\ln2\) (because \(4\ln2-\ln2 = 3\ln2 eq0\)), so the two - sided limit \(\lim_{x ightarrow2}f(x)\) does not exist? But wait, that can't be right. Wait, maybe I misread the problem. Wait, let's check the problem again.
Wait, the function is \(f(x)=\begin{cases}\ln x&0 < x\leq2\\x^{2}\ln2&2 < x\leq4\end{cases}\)
Wait, left - hand limit: \(x ightarrow2^{-}\), \(f(x)=\ln x\), so \(\lim_{x ightarrow2^{-}}f(x)=\ln2\).
Right - hand limit: \(x ightarrow2^{+}\), \(f(x)=x^{2}\ln2\), so \(\lim_{x ightarrow2^{+}}f(x)=2^{2}\ln2 = 4\ln2\).
Since \(\ln2 eq4\ln2\) (because \(4\ln2=\ln2^{4}=\ln16\) and \(\ln2\) is not equal to \(\ln16\)), the left - hand limit and the right - hand limit are not equal. Therefore, the limit \(\lim_{x ightarrow2}f(x)\) does not exist? But wait, that contradicts my initial thought. Wait, no, let's check the answer options. The options are (A) \(\ln2\), (B) \(\ln8\), (C) \(\ln16\), (D) \(4\), (E) nonexistent.
Wait, but maybe I made a mistake in calculating the right - hand limit. Wait, no, the function for \(2 < x\leq4\) is \(x^{2}\ln2\). So when \(x\) approaches \(2\) from the right, we substitute \(x = 2\) into \(x^{2}\ln2\), getting \(4\ln2\). The left - hand limit is \(\ln2\). Since \(4\ln2 eq\ln2\), the two - sided limit does not exist.

Wait, but let's re - evaluate:
Left - hand limit (\(x ightarrow2^{-}\)): \(f(x)=\ln x\), so \(\lim_{x ightarrow2^{-}}f(x)=\ln2\).
Right - hand limit (\(x ightarrow2^{+}\)): \(f(x)=x^{2}\ln2\), so \(\lim_{x ightarrow2^{+}}f(x)=2^{2}\ln2 = 4\ln2\).
Since \(\lim_{x ightarrow2^{-}}f(x) eq\lim_{x ightarrow2^{+}}f(x)\), the limit \(\lim_{x ightarrow2}f(x)\) does not exist.

Answer:

E. nonexistent