JEE MAIN - Mathematics (2020 - 5th September Evening Slot - No. 6)
$$\mathop {\lim }\limits_{x \to 0} {{x\left( {{e^{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)/x}} - 1} \right)} \over {\sqrt {1 + {x^2} + {x^4}} - 1}}$$
is equal to 0.
is equal to $$\sqrt e $$.
is equal to 1.
does not exist.
Explanation
$$\mathop {\lim }\limits_{x \to 0} {{x\left( {{e^{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)/x}} - 1} \right)} \over {\sqrt {1 + {x^2} + {x^4}} - 1}}$$
= $$\mathop {\lim }\limits_{x \to 0} {{x\left[ {{e^{{{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {x\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}$$
= $$\mathop {\lim }\limits_{x \to 0} {{x\left[ {{e^{{{\left( {1 + {x^2} + {x^4}} \right) - 1} \over {x\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {\left( {1 + {x^2} + {x^4} - 1} \right)}}$$
= $$\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{{x^2} + {x^4}} \over {x\left( {\sqrt {1 + 0 + 0} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + 0 + 0} + 1} \right)} \over {x + {x^3}}}$$
= $$2\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{{x^2} + {x^4}} \over {2x}}}} - 1} \right]} \over {x + {x^3}}}$$
= $$2\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{x + {x^3}} \over 2}}} - 1} \right]} \over {{{x + {x^3}} \over 2} \times 2}}$$
= $$2 \times {1 \over 2} \times 1$$
= 1
Note : 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
= $$\mathop {\lim }\limits_{x \to 0} {{x\left[ {{e^{{{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {x\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}$$
= $$\mathop {\lim }\limits_{x \to 0} {{x\left[ {{e^{{{\left( {1 + {x^2} + {x^4}} \right) - 1} \over {x\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {\left( {1 + {x^2} + {x^4} - 1} \right)}}$$
= $$\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{{x^2} + {x^4}} \over {x\left( {\sqrt {1 + 0 + 0} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + 0 + 0} + 1} \right)} \over {x + {x^3}}}$$
= $$2\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{{x^2} + {x^4}} \over {2x}}}} - 1} \right]} \over {x + {x^3}}}$$
= $$2\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{x + {x^3}} \over 2}}} - 1} \right]} \over {{{x + {x^3}} \over 2} \times 2}}$$
= $$2 \times {1 \over 2} \times 1$$
= 1
Note : 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
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