lim _{x rightarrow 2}left(frac{5^x+5^{3-x}-30 | vedclass

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Question

(C) Let $L = \lim _{x \rightarrow 2}\left(\frac{5^x+5^{3-x}-30}{5^{3-x}-5^{\frac{x}{2}}}\right)$.Let $t = 5^{\frac{x}{2}}$. As $x \rightarrow 2$,$t \r

Vedclass · Accepted Answer

$\frac{-8}{3}$

Vedclass · Answer

(C) Let $L = \lim _{x ightarrow 2}\left(\frac{5^x+5^{3-x}-30}{5^{3-x}-5^{\frac{x}{2}}}ight)$.Let $t = 5^{\frac{x}{2}}$. As $x ightarrow 2$,$t ightarrow 5^1 = 5$.Then $5^x = t^2$ and $5^{3-x} = \frac{125}{5^x} = \frac{125}{t^2}$.Substituting these into the limit:$L = \lim _{t ightarrow 5} \frac{t^2 + \frac{125}{t^2} - 30}{\frac{125}{t^2} - t} = \lim _{t ightarrow 5} \frac{t^4 - 30t^2 + 125}{125 - t^3}$.Factorizing the numerator: $t^4 - 30t^2 + 125 = (t^2 - 25)(t^2 - 5) = (t-5)(t+5)(t^2-5)$.Factorizing the denominator: $125 - t^3 = (5-t)(25 + 5t + t^2) = -(t-5)(25 + 5t + t^2)$.Thus,$L = \lim _{t ightarrow 5} \frac{(t-5)(t+5)(t^2-5)}{-(t-5)(25 + 5t + t^2)} = \lim _{t ightarrow 5} \frac{-(t+5)(t^2-5)}{25 + 5t + t^2}$.Substituting $t = 5$:$L = \frac{-(5+5)(25-5)}{25 + 5(5) + 25} = \frac{-10 	imes 20}{75} = \frac{-200}{75} = \frac{-8}{3}$.

$\lim _{x \rightarrow 2}\left(\frac{5^x+5^{3-x}-30}{5^{3-x}-5^{\frac{x}{2}}}\right)=$

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