If $f(x) = \begin{cases} x e^{-\left( \frac{1}{|x|} + \frac{1}{x} \right)}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ then $f(x)$ is Options: (A) discontinuous every where (B) continuous as well as differentiable for all x (C) continuous for all x but not differentiable at x = 0 (D) neither differentiable nor continuous at x = 0

### Analyzing the Setup The function is defined as $f(x) = x e^{-\left( \frac{1}{|x|} + \frac{1}{x} \right)}$ for $x \neq 0$, with the specific condition $f(0) = 0$. To analyze this function, we must resolve the absolute value term $|x|$, which creates two distinct domains: the positive realm ($x > 0$) and the negative realm ($x 0$, we have $|x| = x$. Substituting this into the function, the exponent simplifies as follows: $$f(x) = x e^{-\left( \frac{1}{x} + \frac{1}{x} \right)} = x e^{-\frac{2}{x}}$$ For $x < 0$, we have $|x| = -x$. Substituting this into the function, the exponent becomes: $$f(x) = x e^{-\left( \frac{1}{-x} + \frac{1}{x} \right)} = x e^0 = x$$ This reveals that for all negative values of $x$, the function behaves as a simple linear identity function, $f(x) = x$. ### The Bridge at the Origin To verify continuity at $x=0$, we compare the Left Hand Limit (L.H.L.) and the Right Hand Limit (R.H.L.). For the L.H.L., we approach from the negative side: $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x) = 0$$ For the R.H.L., we approach from the positive side: $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x e^{-\frac{2}{x}}$$ As $x \to 0^+$, the term $\frac{2}{x} \to \infty$, which implies $e^{-\frac{2}{x}} \to 0$. Thus, the limit is $0 \cdot 0 = 0$. Since the L.H.L., R.H.L., and $f(0)$ are all equal, the function is **continuous** at $x=0$. ### The Sharp Turn To test for differentiability, we calculate the Left Hand Derivative (L.H.D.) and the Right Hand Derivative (R.H.D.) at $x=0$ using the definition of the derivative. The L.H.D. is calculated as: $$\lim_{h \to 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{-h - 0}{-h} = 1$$ The R.H.D. is calculated as: $$\lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h e^{-\frac{2}{h}} - 0}{h} = \lim_{h \to 0^+} e^{-\frac{2}{h}} = 0$$ ### Conclusion Because the L.H.D. ($1$) does not equal the R.H.D. ($0$), the function possesses a sharp corner at the origin. Therefore, the function is **continuous everywhere** but **not differentiable at $x=0$**.

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f (x) = {x e^{- (\frac{1}{∣ x ∣} + \frac{1}{x})}, 0, x \neq = 0 x = 0$ then $f (x)$ is

Q: Wait, why does the exponent become zero for $x < 0$? Did I miss a sign?

Great catch! When $x < 0$, the absolute value $|x|$ is defined as $-x$. So, the term $\frac{1}{|x|}$ becomes $\frac{1}{-x}$. When you add that to $\frac{1}{x}$, you get $-\frac{1}{x} + \frac{1}{x} = 0$. Since $e^0 = 1$, the function simply collapses into $f(x) = x$ for all negative values. It's a beautiful simplification that hides in plain sight!

Q: Why is the Right Hand Limit zero? Isn't $0 \cdot \infty$ an indeterminate form?

You are absolutely right to be cautious! However, in the expression $x e^{-\frac{2}{x}}$, as $x \to 0^+$, the exponential term $e^{-\frac{2}{x}}$ decays to zero much faster than $x$ approaches zero. You can think of it as $\frac{x}{e^{2/x}}$. As $x \to 0^+$, the denominator grows exponentially to infinity, forcing the entire fraction to zero. It's a classic case of exponential dominance.

Q: Why can't I just differentiate the whole thing using the chain rule and plug in $x=0$?

That is a very common trap! The derivative formula only works if the function is already known to be differentiable. At $x=0$, the function is defined piecewise. The derivative at a point of transition must be checked using the first-principles definition (the limit of the difference quotient) because the algebraic formula might not be valid exactly at the junction.

Q: If the L.H.D. is 1 and the R.H.D. is 0, does that mean the function is discontinuous?

Not at all! Continuity and differentiability are different beasts. Continuity only asks if the graph is connected (no jumps). Differentiability asks if the graph is 'smooth' (no sharp corners). Here, the graph is connected at $(0,0)$, but it arrives there with a slope of 1 from the left and a slope of 0 from the right, creating a sharp corner.

Q: Is there a visual way to think about this?

Imagine a path. For $x < 0$, you are walking along the line $y=x$, which has a constant slope of 1. As you approach the origin, you suddenly transition into a curve that flattens out to a slope of 0 as it enters the origin from the right. Because you change your 'direction' (slope) instantly at the origin, you have a sharp corner, even though your feet never leave the ground!

Select Answer:

Visualized Solution

Understanding the Piecewise Function $f (x)$

Given function: $f (x) = {x e^{- (\frac{1}{∣ x ∣} + \frac{1}{x})}, 0, x \neq = 0 x = 0$
We need to analyze the behavior of $∣ x ∣$ to simplify the expression for $x > 0$ and $x < 0$ .

Raw Setup for $x > 0$

For $x > 0$ , $∣ x ∣ = x$ .
Substituting this: $f (x) = x e^{- (\frac{1}{x} + \frac{1}{x})}$

Simplifying for $x > 0$

Simplifying the exponent: $\frac{1}{x} + \frac{1}{x} = \frac{2}{x}$
$f (x) = x e^{- \frac{2}{x}}$ for $x > 0$

Raw Setup for $x < 0$

For $x < 0$ , $∣ x ∣ = - x$ .
Substituting this: $f (x) = x e^{- (\frac{1}{- x} + \frac{1}{x})}$

Simplifying for $x < 0$

Simplifying the exponent: $- \frac{1}{x} + \frac{1}{x} = 0$
$f (x) = x e^{0} = x$ for $x < 0$

Checking Left Hand Limit (L.H.L.)

Left Hand Limit (L.H.L.) at $x = 0$ :
$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} x = 0$

Checking Right Hand Limit (R.H.L.)

Right Hand Limit (R.H.L.) at $x = 0$ :
$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} x e^{- \frac{2}{x}}$
As $x \to 0^{+}$ , $\frac{2}{x} \to \infty$ , so $e^{- \frac{2}{x}} \to 0$ .
R.H.L. $= 0 \cdot 0 = 0$

Conclusion on Continuity

L.H.L. $=$ R.H.L. $= f (0) = 0$
Therefore, $f (x)$ is continuous for all $x$ .

Checking Left Hand Derivative (L.H.D.)

Left Hand Derivative (L.H.D.) at $x = 0$ :
$lim_{h \to 0} \frac{f ( 0 - h ) - f ( 0 )}{- h}$
$= lim_{h \to 0} \frac{- h - 0}{- h} = 1$

Checking Right Hand Derivative (R.H.D.)

Right Hand Derivative (R.H.D.) at $x = 0$ :
$lim_{h \to 0} \frac{f ( 0 + h ) - f ( 0 )}{h}$
$= lim_{h \to 0} \frac{h e ^{- \frac{2}{h}} - 0}{h} = lim_{h \to 0} e^{- \frac{2}{h}} = 0$

Final Verdict on Differentiability

L.H.D. ( $1$ ) $\neq =$ R.H.D. ( $0$ )
The function has a sharp corner at $x = 0$ .
Final Result: Continuous for all $x$ but not differentiable at $x = 0$ .

00:00 / 00:00

The Sigma Insight: Relationship Between Continuity and Differentiability

Solution Diagram

Analyzing the Setup

The function is defined as

f (x) = x e^{- (\frac{1}{∣ x ∣} + \frac{1}{x})}

for $x eq 0$, with the specific condition

f (0) = 0

To analyze this function, we must resolve the absolute value term

∣ x ∣

, which creates two distinct domains: the positive realm (

x > 0

) and the negative realm (

x < 0

The Two Worlds of $x$

For

x > 0

, we have

∣ x ∣ = x

. Substituting this into the function, the exponent simplifies as follows:

f (x) = x e^{- (\frac{1}{x} + \frac{1}{x})} = x e^{- \frac{2}{x}}

For

x < 0

, we have

∣ x ∣ = - x

. Substituting this into the function, the exponent becomes:

f (x) = x e^{- (\frac{1}{- x} + \frac{1}{x})} = x e^{0} = x

This reveals that for all negative values of

x

, the function behaves as a simple linear identity function,

f (x) = x

The Bridge at the Origin

To verify continuity at

x = 0

, we compare the Left Hand Limit (L.H.L.) and the Right Hand Limit (R.H.L.).

For the L.H.L., we approach from the negative side:

x \to 0^{-} lim f (x) = x \to 0^{-} lim (x) = 0

For the R.H.L., we approach from the positive side:

x \to 0^{+} lim f (x) = x \to 0^{+} lim x e^{- \frac{2}{x}}

x \to 0^{+}

, the term

\frac{2}{x} \to \infty

, which implies

e^{- \frac{2}{x}} \to 0

. Thus, the limit is

0 \cdot 0 = 0

. Since the L.H.L., R.H.L., and

f (0)

are all equal, the function is continuous at

x = 0

The Sharp Turn

To test for differentiability, we calculate the Left Hand Derivative (L.H.D.) and the Right Hand Derivative (R.H.D.) at

x = 0

using the definition of the derivative.

The L.H.D. is calculated as:

h \to 0^{+} lim \frac{f ( 0 - h ) - f ( 0 )}{- h} = h \to 0^{+} lim \frac{- h - 0}{- h} = 1

The R.H.D. is calculated as:

h \to 0^{+} lim \frac{f ( 0 + h ) - f ( 0 )}{h} = h \to 0^{+} lim \frac{h e ^{- \frac{2}{h}} - 0}{h} = h \to 0^{+} lim e^{- \frac{2}{h}} = 0

Conclusion

Because the L.H.D. (

1

) does not equal the R.H.D. (

0

), the function possesses a sharp corner at the origin.

Therefore, the function is continuous everywhere but not differentiable at $x = 0$ .

Common Traps & FAQs

Wait, why does the exponent become zero for

x < 0

? Did I miss a sign?

Great catch! When

x < 0

, the absolute value

∣ x ∣

is defined as

- x

. So, the term

\frac{1}{∣ x ∣}

becomes

\frac{1}{- x}

. When you add that to

\frac{1}{x}

, you get

- \frac{1}{x} + \frac{1}{x} = 0

. Since

e^{0} = 1

, the function simply collapses into

f (x) = x

for all negative values. It's a beautiful simplification that hides in plain sight!

Why is the Right Hand Limit zero? Isn't

0 \cdot \infty

an indeterminate form?

You are absolutely right to be cautious! However, in the expression

x e^{- \frac{2}{x}}

, as

x \to 0^{+}

, the exponential term

e^{- \frac{2}{x}}

decays to zero much faster than

x

approaches zero. You can think of it as

\frac{x}{e ^{2/ x}}

. As

x \to 0^{+}

, the denominator grows exponentially to infinity, forcing the entire fraction to zero. It's a classic case of exponential dominance.

Why can't I just differentiate the whole thing using the chain rule and plug in

x = 0

That is a very common trap! The derivative formula only works if the function is already known to be differentiable. At

x = 0

, the function is defined piecewise. The derivative at a point of transition must be checked using the first-principles definition (the limit of the difference quotient) because the algebraic formula might not be valid exactly at the junction.

If the L.H.D. is 1 and the R.H.D. is 0, does that mean the function is discontinuous?

Not at all! Continuity and differentiability are different beasts. Continuity only asks if the graph is connected (no jumps). Differentiability asks if the graph is 'smooth' (no sharp corners). Here, the graph is connected at

(0, 0)

, but it arrives there with a slope of 1 from the left and a slope of 0 from the right, creating a sharp corner.

Is there a visual way to think about this?

Imagine a path. For

x < 0

, you are walking along the line

y = x

, which has a constant slope of 1. As you approach the origin, you suddenly transition into a curve that flattens out to a slope of 0 as it enters the origin from the right. Because you change your 'direction' (slope) instantly at the origin, you have a sharp corner, even though your feet never leave the ground!

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f (x) = {x e^{- (\frac{1}{∣ x ∣} + \frac{1}{x})}, 0, x \neq = 0 x = 0$ then $f (x)$ is

Select Answer:

Visualized Solution

Understanding the Piecewise Function $f (x)$

Raw Setup for $x > 0$

Simplifying for $x > 0$

Raw Setup for $x < 0$

Simplifying for $x < 0$

Checking Left Hand Limit (L.H.L.)

Checking Right Hand Limit (R.H.L.)

Conclusion on Continuity

Checking Left Hand Derivative (L.H.D.)

Checking Right Hand Derivative (R.H.D.)

Final Verdict on Differentiability

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Two Worlds of $x$

The Bridge at the Origin

The Sharp Turn

Conclusion

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)={xe−(∣x∣1​+x1​),0,​x=0x=0​ then f(x) is

Select Answer:

Visualized Solution

Understanding the Piecewise Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

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The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

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The Bridge at the Origin

The Sharp Turn

Conclusion

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