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Limits Continuity and Differentiability question

2019 · Shift 2 · Q24
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  5. /2019 · Shift 2 · Q24

Limits Continuity and Differentiability question

2019 · Shift 2 · Q24

JEE AdvancedMathematicsLimits Continuity and DifferentiabilityMultiple correct+4 / −1
Let f : R be a function. We say that f has PROPERTY 1 if lim⁡h→0f(h)−f(0)∣h∣\mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over {\sqrt {|h|} }}h→0lim​∣h∣​f(h)−f(0)​ exists and is finite, and PROPERTY 2 if lim⁡h→0f(h)−f(0)h2\mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over {{h^2}}}h→0lim​h2f(h)−f(0)​ exists and is finite. Then which of the following options is/are correct?
  1. A
    f(x) = sin x has PROPERTY 2
  2. B
    f(x) = x2/3 has PROPERTY 1
  3. C
    f(x) = |x| has PROPERTY 1
  4. D
    f(x) = x|x| has PROPERTY 2
View written solutionFree

Correct answer: C, B

We check each function using the given definitions.

Given properties

A function fff has:

  1. PROPERTY 1 if
lim⁡h→0f(h)−f(0)∣h∣\lim_{h\to 0}\frac{f(h)-f(0)}{\sqrt{|h|}}h→0lim​∣h∣​f(h)−f(0)​

exists and is finite.

  1. PROPERTY 2 if
lim⁡h→0f(h)−f(0)h2\lim_{h\to 0}\frac{f(h)-f(0)}{h^2}h→0lim​h2f(h)−f(0)​

exists and is finite.

We now test each option.

Option A: f(x)=sin⁡xf(x)=\sin xf(x)=sinx has PROPERTY 2

Here,

f(0)=sin⁡0=0.f(0)=\sin 0=0.f(0)=sin0=0.

So we need

lim⁡h→0sin⁡hh2.\lim_{h\to 0}\frac{\sin h}{h^2}.h→0lim​h2sinh​.

Rewrite as

sin⁡hh2=(sin⁡hh)1h.\frac{\sin h}{h^2}=\left(\frac{\sin h}{h}\right)\frac{1}{h}.h2sinh​=(hsinh​)h1​.

As h→0h\to 0h→0,

sin⁡hh→1,\frac{\sin h}{h}\to 1,hsinh​→1,

but

1h\frac{1}{h}h1​

does not have a finite limit. Hence the limit does not exist finitely.

So Option A is false.

Option B: f(x)=x2/3f(x)=x^{2/3}f(x)=x2/3 has PROPERTY 1

Here,

f(0)=0.f(0)=0.f(0)=0.

We need

lim⁡h→0h2/3∣h∣.\lim_{h\to 0}\frac{h^{2/3}}{\sqrt{|h|}}.h→0lim​∣h∣​h2/3​.

Since h2/3=∣h∣2/3h^{2/3}=|h|^{2/3}h2/3=∣h∣2/3 for real hhh,

h2/3∣h∣=∣h∣2/3−1/2=∣h∣1/6.\frac{h^{2/3}}{\sqrt{|h|}}=|h|^{2/3-1/2}=|h|^{1/6}.∣h∣​h2/3​=∣h∣2/3−1/2=∣h∣1/6.

Now as h→0h\to 0h→0,

∣h∣1/6→0.|h|^{1/6}\to 0.∣h∣1/6→0.

The limit exists and is finite.

So Option B is true.

Option C: f(x)=∣x∣f(x)=|x|f(x)=∣x∣ has PROPERTY 1

Here,

f(0)=0.f(0)=0.f(0)=0.

We need

lim⁡h→0∣h∣∣h∣=lim⁡h→0∣h∣.\lim_{h\to 0}\frac{|h|}{\sqrt{|h|}}=\lim_{h\to 0}\sqrt{|h|}.h→0lim​∣h∣​∣h∣​=h→0lim​∣h∣​.

Since

∣h∣→0,\sqrt{|h|}\to 0,∣h∣​→0,

the limit exists and is finite.

So Option C is true.

Option D: f(x)=x∣x∣f(x)=x|x|f(x)=x∣x∣ has PROPERTY 2

Here,

f(0)=0.f(0)=0.f(0)=0.

We need

lim⁡h→0h∣h∣h2.\lim_{h\to 0}\frac{h|h|}{h^2}.h→0lim​h2h∣h∣​.

For h≠0h\neq 0h=0,

h∣h∣h2=∣h∣h={1,h>0,−1,h<0.\frac{h|h|}{h^2}=\frac{|h|}{h}= \begin{cases} 1, & h>0,\\ -1, & h<0. \end{cases}h2h∣h∣​=h∣h∣​={1,−1,​h>0,h<0.​

Thus the right-hand limit is 111 and the left-hand limit is −1-1−1. So the limit does not exist.

Hence Option D is false.

Final conclusion

The correct options are:

B, C\boxed{B,\ C}B, C​

Comparison with stored correct answer

Stored correct answer: C,BC, BC,B

This matches our derived answer (order does not matter).

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