limlimits_{x rightarrow 2} frac{3^{x}+3^{3-x} | vedclass

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Question

(B) Let $L = \lim _{x \rightarrow 2} \frac{3^{x}+3^{3-x}-12}{3^{-x / 2}-3^{1-x}}$.Substitute $t = 3^{x/2}$. As $x \rightarrow 2$,$t \rightarrow 3^{2/2

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) Let $L = \lim _{x ightarrow 2} \frac{3^{x}+3^{3-x}-12}{3^{-x / 2}-3^{1-x}}$.Substitute $t = 3^{x/2}$. As $x ightarrow 2$,$t ightarrow 3^{2/2} = 3$.Then $3^x = t^2$ and $3^{3-x} = \frac{27}{3^x} = \frac{27}{t^2}$.Also,$3^{-x/2} = \frac{1}{t}$ and $3^{1-x} = \frac{3}{3^x} = \frac{3}{t^2}$.The expression becomes:$L = \lim _{t ightarrow 3} \frac{t^2 + \frac{27}{t^2} - 12}{\frac{1}{t} - \frac{3}{t^2}} = \lim _{t ightarrow 3} \frac{\frac{t^4 - 12t^2 + 27}{t^2}}{\frac{t - 3}{t^2}} = \lim _{t ightarrow 3} \frac{t^4 - 12t^2 + 27}{t - 3}$.Factor the numerator: $t^4 - 12t^2 + 27 = (t^2 - 9)(t^2 - 3) = (t - 3)(t + 3)(t^2 - 3)$.$L = \lim _{t ightarrow 3} \frac{(t - 3)(t + 3)(t^2 - 3)}{t - 3} = \lim _{t ightarrow 3} (t + 3)(t^2 - 3)$.Substituting $t = 3$: $L = (3 + 3)(3^2 - 3) = 6 	imes (9 - 3) = 6 	imes 6 = 36$.

$\lim\limits_{x \rightarrow 2} \frac{3^{x}+3^{3-x}-12}{3^{-x / 2}-3^{1-x}}$ is equal to

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