Medium MCQ +4 / -1 PYQ · JEE Mains 2020

$$\mathop {\lim }\limits_{x \to 0} {{x\left( {{e^{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)/x}} - 1} \right)} \over {\sqrt {1 + {x^2} + {x^4}} - 1}}$$

  1. A is equal to 0.
  2. B is equal to $\sqrt e$.
  3. C is equal to 1. Correct answer
  4. D does not exist.

Solution

$$\mathop {\lim }\limits_{x \to 0} {{x\left( {{e^{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)/x}} - 1} \right)} \over {\sqrt {1 + {x^2} + {x^4}} - 1}}$$ <br><br>= $$\mathop {\lim }\limits_{x \to 0} {{x\left[ {{e^{{{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {x\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}$$ <br><br>= $$\mathop {\lim }\limits_{x \to 0} {{x\left[ {{e^{{{\left( {1 + {x^2} + {x^4}} \right) - 1} \over {x\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {\left( {1 + {x^2} + {x^4} - 1} \right)}}$$ <br><br>= $$\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{{x^2} + {x^4}} \over {x\left( {\sqrt {1 + 0 + 0} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + 0 + 0} + 1} \right)} \over {x + {x^3}}}$$ <br><br>= $$2\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{{x^2} + {x^4}} \over {2x}}}} - 1} \right]} \over {x + {x^3}}}$$ <br><br>= $$2\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{x + {x^3}} \over 2}}} - 1} \right]} \over {{{x + {x^3}} \over 2} \times 2}}$$ <br><br>= $2 \times {1 \over 2} \times 1$ <br><br>= 1 <br><br><b>Note :</b> As from formula, $$\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{x + {x^3}} \over 2}}} - 1} \right]} \over {{{x + {x^3}} \over 2}}}$$ = 1

About this question

Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results

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