Let f(x)={[ln(x)," for "0 2]:} Is f continuous at x=2 ? Choose 1 answer: (A) Yes (B) No

Check Conditions: To determine if the function f(x) is continuous at x=2, we need to check if the following three conditions are met: 1. f(2) is defined. 2. The limit of f(x) as x approaches 2 from the left (lim x->2^- f(x)) exists. 3. The limit of f(x) as x approaches 2 from the right (lim x->2^+ f(x)) exists and is equal to the limit from the left and to f(2). First, we will find f(2) using the definition of the function for 0 2^- f(x) = ln(2).Right Limit: Now, we need to find the limit of f(x) as x approaches 2 from the right. For x > 2, the function is defined as x^2 * ln(2). We need to evaluate this limit as x approaches 2. lim x->2^+ f(x) = (2^2) * ln(2) = 4 * ln(2).Comparison: We compare the limits from the left and the right and the value of the function at x=2. Since lim x->2^- f(x) = ln(2) and lim x->2^+ f(x) = 4 * ln(2), the limits are not equal. Therefore, the function is not continuous at x=2 because the limit from the right does not equal the limit from the left nor the value of the function at x=2.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

$Let f(x)={[ln(x)," for "0 < x <= 2],[x^(2)ln(2)," for "x > 2]:} Is f continuous at x=2 ? Choose 1 answer: (A) Yes (B) No$

Let $\newline$ $f(x)=\left\{\begin{array}{ll}\ln (x) & \text { for } 0<x 2="" \leq="" \\="" x^{2}="" \ln="" (2)="" &="" \text="" {="" for="" }="" x="">2\end{array}\right.$ $\newline$ Is $f$ continuous at $x=2$ ? $\newline$ Choose $1$ answer: $\newline$ (A) Yes $\newline$ (B) $\mathrm{No}$

View step-by-step help

Home

Math Problems

Calculus

Intermediate Value Theorem

Full solution

Q. Let

\newline

f(x)=\left\{\begin{array}{ll}\ln (x) & \text { for } 0<x \leq 2 \\ x^{2} \ln (2) & \text { for } x>2\end{array}\right.

\newline

f

continuous at

x=2

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B)

\mathrm{No}

Check Conditions: To determine if the function $f(x)$ is continuous at $x=2$ , we need to check if the following three conditions are met: $\newline$ $1$ . $f(2)$ is defined. $\newline$ $2$ . The limit of $f(x)$ as $x$ approaches $2$ from the left ( $\lim_{x\to 2^-} f(x)$ ) exists. $\newline$ $3$ . The limit of $f(x)$ as $x$ approaches $2$ from the right ( $x=2$ $0$ ) exists and is equal to the limit from the left and to $f(2)$ . $\newline$ First, we will find $f(2)$ using the definition of the function for $x=2$ $3$ .
Find $f(2)$ : Substitute $x = 2$ into the first part of the function definition to find $f(2)$ . $\newline$ $f(2) = \ln(2)$ . $\newline$ This value is defined, so the first condition for continuity is met.
Left Limit: Next, we need to find the limit of $f(x)$ as $x$ approaches $2$ from the left. $\newline$ Since the function for 0 < x \leq 2 is $\ln(x)$ , the limit as $x$ approaches $2$ from the left is simply the value of the function at $x=2$ . $\newline$ $\lim_{x\to 2^-} f(x) = \ln(2)$ .
Right Limit: Now, we need to find the limit of $f(x)$ as $x$ approaches $2$ from the right. $\newline$ For x > 2, the function is defined as $x^2 \cdot \ln(2)$ . We need to evaluate this limit as $x$ approaches $2$ . $\newline$ $\lim_{x\to 2^+} f(x) = (2^2) \cdot \ln(2) = 4 \cdot \ln(2)$ .
Comparison: We compare the limits from the left and the right and the value of the function at $x=2$ . $\newline$ Since $\lim_{x\to 2^-} f(x) = \ln(2)$ and $\lim_{x\to 2^+} f(x) = 4 \times \ln(2)$ , the limits are not equal. $\newline$ Therefore, the function is not continuous at $x=2$ because the limit from the right does not equal the limit from the left nor the value of the function at $x=2$ .