The value of mathop {lim }limits_{x to 2} fra | vedclass

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Question

(C) Given the limit: $\mathop {\lim }\limits_{x 	o 2} \frac{{{3^{x/2}} - 3}}{{{3^x} - 9}}$We can rewrite the denominator as a difference of squares:

Vedclass · Accepted Answer

$1/6$

Vedclass · Answer

(C) Given the limit: $\mathop {\lim }\limits_{x 	o 2} \frac{{{3^{x/2}} - 3}}{{{3^x} - 9}}$We can rewrite the denominator as a difference of squares: ${3^x} - 9 = {({3^{x/2}})^2} - {3^2} = ({3^{x/2}} - 3)({3^{x/2}} + 3)$Substituting this into the limit expression:$\mathop {\lim }\limits_{x 	o 2} \frac{{{3^{x/2}} - 3}}{({3^{x/2}} - 3)({3^{x/2}} + 3)}$Canceling the common term $({3^{x/2}} - 3)$:$\mathop {\lim }\limits_{x 	o 2} \frac{1}{{{3^{x/2}} + 3}}$Evaluating the limit as $x 	o 2$:$\frac{1}{{{3^{2/2}} + 3}} = \frac{1}{{{3^1} + 3}} = \frac{1}{3 + 3} = \frac{1}{6}$

The value of $\mathop {\lim }\limits_{x \to 2} \frac{{{3^{x/2}} - 3}}{{{3^x} - 9}}$ is

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