JEE MAIN - Mathematics (2020 - 7th January Morning Slot - No. 8)
$$\mathop {\lim }\limits_{x \to 2} {{{3^x} + {3^{3 - x}} - 12} \over {{3^{ - x/2}} - {3^{1 - x}}}}$$ is equal to_______.
Answer
36
Explanation
$$\mathop {\lim }\limits_{x \to 2} {{{3^x} + {3^{3 - x}} - 12} \over {{3^{ - x/2}} - {3^{1 - x}}}}$$
let 3x/2 = t
= $$\mathop {\lim }\limits_{t \to 3} {{{t^2} + {{27} \over {{t^2}}} - 12} \over {{1 \over t} - {3 \over {{t^2}}}}}$$
= $$\mathop {\lim }\limits_{t \to 3} {{\left( {{t^2} - 9} \right)\left( {{t^2} - 3} \right)} \over {t - 3}}$$
= $$\mathop {\lim }\limits_{t \to 3} \left( {t + 3} \right)\left( {{t^2} - 3} \right)$$
= 6 $$ \times $$ 6
= 36
let 3x/2 = t
= $$\mathop {\lim }\limits_{t \to 3} {{{t^2} + {{27} \over {{t^2}}} - 12} \over {{1 \over t} - {3 \over {{t^2}}}}}$$
= $$\mathop {\lim }\limits_{t \to 3} {{\left( {{t^2} - 9} \right)\left( {{t^2} - 3} \right)} \over {t - 3}}$$
= $$\mathop {\lim }\limits_{t \to 3} \left( {t + 3} \right)\left( {{t^2} - 3} \right)$$
= 6 $$ \times $$ 6
= 36
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