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Limits Continuity and Differentiability question

2020 · 5 Sep · Shift 2 · Q24
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  5. /2020 · 5 Sep · Shift 2 · Q24

Limits Continuity and Differentiability question

2020 · 5 Sep · Shift 2 · Q24

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
lim⁡x→0x(e(1+x2+x4−1)/x−1)1+x2+x4−1\mathop {\lim }\limits_{x \to 0} {{x\left( {{e^{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)/x}} - 1} \right)} \over {\sqrt {1 + {x^2} + {x^4}} - 1}}x→0lim​1+x2+x4​−1x(e(1+x2+x4​−1)/x−1)​
  1. A
    is equal to 0.
  2. B
    is equal to e\sqrt ee​.
  3. C
    is equal to 1.
  4. D
    does not exist.
View written solutionFree

Correct answer: C

  1. Given limit

We need to evaluate

L=lim⁡x→0x(e(1+x2+x4−1)/x−1)1+x2+x4−1.L=\lim_{x\to 0} \frac{x\left(e^{\left(\sqrt{1+x^2+x^4}-1\right)/x}-1\right)}{\sqrt{1+x^2+x^4}-1}.L=x→0lim​1+x2+x4​−1x(e(1+x2+x4​−1)/x−1)​.
  1. Introduce a substitution

Let

y=1+x2+x4−1x.y=\frac{\sqrt{1+x^2+x^4}-1}{x}.y=x1+x2+x4​−1​.

Then the expression becomes

L=lim⁡x→0ey−1y,L=\lim_{x\to 0} \frac{e^y-1}{y},L=x→0lim​yey−1​,

because

x(ey−1)1+x2+x4−1=ey−1(1+x2+x4−1)/x=ey−1y.\frac{x(e^y-1)}{\sqrt{1+x^2+x^4}-1} =\frac{e^y-1}{\left(\sqrt{1+x^2+x^4}-1\right)/x} =\frac{e^y-1}{y}.1+x2+x4​−1x(ey−1)​=(1+x2+x4​−1)/xey−1​=yey−1​.

So now we only need to find lim⁡x→0y\lim_{x\to 0} ylimx→0​y.

  1. Evaluate yyy as x→0x\to 0x→0

We have

y=1+x2+x4−1x.y=\frac{\sqrt{1+x^2+x^4}-1}{x}.y=x1+x2+x4​−1​.

Rationalize the numerator:

y=(1+x2+x4)−1x(1+x2+x4+1)=x2+x4x(1+x2+x4+1).y=\frac{(1+x^2+x^4)-1}{x\left(\sqrt{1+x^2+x^4}+1\right)} =\frac{x^2+x^4}{x\left(\sqrt{1+x^2+x^4}+1\right)}.y=x(1+x2+x4​+1)(1+x2+x4)−1​=x(1+x2+x4​+1)x2+x4​.

Thus,

y=x(1+x2)1+x2+x4+1.y=\frac{x(1+x^2)}{\sqrt{1+x^2+x^4}+1}.y=1+x2+x4​+1x(1+x2)​.

As x→0x\to 0x→0,

y→0⋅(1+0)1+0+0+1=0.y\to \frac{0\cdot (1+0)}{\sqrt{1+0+0}+1}=0.y→1+0+0​+10⋅(1+0)​=0.
  1. Use the standard limit

Now since y→0y\to 0y→0,

L=lim⁡y→0ey−1y=1.L=\lim_{y\to 0}\frac{e^y-1}{y}=1.L=y→0lim​yey−1​=1.
  1. Check options
  • A: 000 ❌
  • B: e\sqrt ee​ ❌
  • C: 111 ✅
  • D: does not exist ❌

Therefore, the correct option is C.

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