Q.Prove that the function defined as f(x)=⎩⎨⎧x⋅e1/x+1e1/x−1,0,if x=0,if x=0 is continuous but not derivable at x=0.
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Continuity Differentiability Relationship
How Continuity and Differentiability Are Related
Two properties describe how "well-behaved" a function is at a point. Continuity means the graph has no break there — you can draw through the point without lifting your pen. Differentiability means the graph is smooth there — it has one definite tangent line, so a well-defined slope f′(a). This concept is about the exact link between the two.
The theorem: If f is differentiable at x=a, then f is continuous at x=a.
Why differentiability forces continuity
If f′(a) exists, then
limx→a(f(x)−f(a))=limx→ax−af(x)−f(a)⋅(x−a)=f′(a)⋅0=0.
So limx→af(x)=f(a), which is exactly continuity at a. A curve that has a tangent cannot also have a jump — a break would send the difference quotient to infinity and the derivative would not exist.
The converse is FALSE
Continuity does not guarantee differentiability. A graph can be unbroken yet still have a sharp corner, and a corner has no single tangent.
The classic counterexample is f(x)=∣x∣ at x=0. It is continuous there (limx→0∣x∣=0=f(0)), but the slope from the left is −1 and from the right is +1. Since these disagree, f′(0) does not exist.
Putting it together
- Differentiable at a ⇒ continuous at a.
- Continuous at a ⇒ differentiable at a.
- Not continuous at a ⇒ not differentiable at a (the contrapositive of the theorem). …
Because 1/x behaves oppositely as x approaches 0 from the left versus the right, the one-sided limits of e^(1/x) -- and hence of f(x) -- must be evaluated separately for continuity, and the same left/right split is needed again for the derivative at 0 …
evaluate the two one-sided limits (as x→1/x behaves oppositely) for continuity and for the derivative
f(x)=x⋅e1/x+1e1/x−1 for x=0, f(0)=0.
Continuity at x=0:
As x→0+, x1→+∞, so e1/x→∞, and e1/x+1e1/x−1→1 (dividing num/denom by e1/x: 1+e−1/x1−e−1/x→1+01−0=1). So f(x)→x⋅1→0 as x→0+.
As x→0−, x1→−∞, so e1/x→0, and e1/x+1e1/x−1→0+10−1=−1. So f(x)→x⋅(−1)→0 as x→0− (since x→0).
Both one-sided limits equal 0=f(0), so f is continuous at x=0.
…
- CBSE 2026Set ANNUAL1 markQ.Write True or False: Every differentiable function is continuous, but the converse is not true.
›Reveal solutionSolution
True - differentiable ⇒ continuous, converse false.
…
- CBSE 2025Set ANNUAL1 markMCQQ.Assertion (A): Every differentiable function is continuous. Reason (R): Every continuous function is differentiable. Choose the correct option:(a) Both A and R are correct and R is the correct explanation of A.(b) Both A and R are correct but R is not the correct explanation of A.(c) A is correct but R is incorrect.(d) Both A and R are incorrect.
›Reveal solutionSolution
Differentiability implies continuity, but the converse is false.
A is TRUE: Every differentiable function is continuous — this is a standard theorem.
…
- CBSE 2023Set 65/3/11 markMCQQ.The value of k for which function f(x)={x2,kx,x≥0x<0 is differentiable at x=0 is :(a) 1(b) 2(c) any real number(d) 0
›Reveal solutionSolution
A piecewise function is differentiable at a point only if it is continuous there and the left and right derivatives match. For f(x) at x=0, continuity forces k to be anything, but matching derivatives forces k=0.
Differentiability is a stronger condition than continuity. For a function to be differentiable at a point, two things must happen: the function must be continuous there, and the derivative must exist (meaning the left-hand and right-hand derivatives must be equal).
Let's check both conditions systematically for f(x) at x=0.
Checking Continuity at x=0
For continuity at x=0, we need:
limx→0−f(x)=limx→0+f(x)=f(0)
-
From the right: As x→0+, we use f(x)=x2, so limx→0+f(x)=02=0.
-
From the left: As x→0−, we use f(x)=kx, so limx→0−f(x)=k⋅0=0.
-
At the point: f(0)=02=0 (since 0≥0, we use the first piece).
All three equal 0 regardless of k, so f is continuous at x=0 for any value of k.
Checking Differentiability at x=0
Now we compute the left-hand derivative (LHD) and right-hand derivative (RHD) using the definition:
f′(0)=limh→0hf(0+h)−f(0)=limh→0hf(h)
Right-hand derivative (h→0+): …
-
- CBSE 2020Set HE8231 markQ.Fill in the blank: Every differentiable function is ______.
›Reveal solutionSolution
Every differentiable function is continuous.
If a function f is differentiable at a point x=a, then
limh→0hf(a+h)−f(a)=f′(a) exists (finite).
Writing f(a+h)−f(a)=hf(a+h)−f(a)⋅h and taking the limit as h→0:
limh→0[f(a+h)−f(a)]=f′(a)⋅0=0,
which means h→0limf(a+h)=f(a) — exactly the condition for continuity at x=a.
…
- CBSE 2019Set ANNUAL1 markQ.If f′(2+)=0 and f′(2−)=0, then is f(x) continuous at x=2?
›Reveal solutionSolution
Existence of a finite one-sided derivative at a point forces one-sided continuity there; since both f′(2+) and f′(2−) exist (and equal 0), f is continuous at x=2.
By definition, f′(2+)=h→0+limhf(2+h)−f(2).
For this limit to exist and be finite (here, equal to 0), the numerator f(2+h)−f(2) must itself tend to 0 as h→0+ — because f(2+h)−f(2)=h⋅hf(2+h)−f(2)→0⋅0=0.
So h→0+limf(2+h)=f(2), i.e. f is right-continuous at x=2.
…
- CBSE 2018Set ANNUAL1 markMCQQ.If f(x) is differentiable at x = a, which of the following statement may be false ?(a) f(x) is continuous at x = a(b) lim x→a f(x) exist(c) lim h→0⁻ [f(a+h) − f(a)]/h = lim h→0⁺ [f(a+h) − f(a)]/h(d) The second derivative of f(x) i.e. f″(x) exist at x = a
›Reveal solutionSolution
Differentiability at a point guarantees continuity and the existence of f′(a), but never guarantees f″(a) exists.
If f is differentiable at x=a, then by definition the limit f′(a)=limh→0hf(a+h)−f(a) exists, which forces:
- (a) f continuous at a — always true (differentiability ⇒ continuity).
- (b) limx→af(x) exists — always true, again because differentiability implies continuity.
- (c) limh→0−hf(a+h)−f(a)=limh→0+hf(a+h)−f(a) — always true, since a two-sided derivative existing means the left-hand and right-hand derivatives are equal. …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.