(d) $\lim _{x \rightarrow-6^{+}} \frac{e^{2 x}}{x+6}$

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∻Solution by Steps∻ ‖ step 1 ⋮ Evaluate the limit from the right side as $x$ approaches $-6$. ‖ ‖ step 2 ⋮ The expression $\frac{e^{2x}}{x+6}$ becomes undefined as $x$ approaches $-6$ from the right, since the denominator approaches $0$. ‖ ‖ step 3 ⋮ As $x$ gets closer to $-6$ from the right, $e^{2x}$ approaches $e^{-12}$, which is a positive constant. ‖ ‖ step 4 ⋮ Since the numerator remains positive and the denominator approaches $0$ from the positive side, the limit tends to $+\infty$. ‖ ∻Answer∻ ⚹ $\lim _{x \rightarrow -6^{+}} \frac{e^{2 x}}{x+6} = +\infty$ ⚹ ∻Key Concept∻ ⚹Limits involving a non-zero constant divided by an expression approaching zero⚹ ∻Explanation∻ ⚹As $x$ approaches $-6$ from the right, the denominator of the fraction $\frac{e^{2x}}{x+6}$ approaches zero, causing the value of the fraction to increase without bound. Hence, the limit is $+\infty$.⚹◊What is the limit of $\frac{e^{2x}}{x+6}$ as $x$ approaches $-6$ from the right side?⍭ Generate me a similar question◊

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