Explained: "I needed to present a method to finding..."

Adison Palmer

Answered question

2022-04-06

I needed to present a method to finding the limit of

lim_{x \to 0} \frac{9^{x} - 5^{x}}{x}

annieljcddj0

Beginner2022-04-07Added 15 answers

Remember that

lim_{y \to 0} \frac{e^{y} - 1}{y} = 1

. Set

y = \log (\frac{9}{5}) \cdot x

and note that

lim_{x \to 0} \frac{9^{x} - 5^{x}}{x} = lim_{x \to 0} (5^{x}) \cdot \frac{{(\frac{9}{5})}^{x} - 1}{x}

= lim_{x \to 0} (5^{x}) \cdot \frac{e^{x \log (\frac{9}{5})} - 1}{x}

= lim_{x \to 0} (\log (\frac{9}{5}) \cdot 5^{x}) \cdot \frac{e^{x \log (\frac{9}{5})} - 1}{x \cdot \log (\frac{9}{5})}

= \log (\frac{9}{5}) \cdot lim_{x \to 0} (5^{x}) lim_{y \to 0} \frac{e^{y} - 1}{y}

= \log (\frac{9}{5}) \cdot 5^{0} \cdot lim_{y \to 0} lim_{y \to 0} \frac{e^{y} - 1}{y}

= \log (\frac{9}{5})

aludirao09au

Beginner2022-04-08Added 11 answers

You could recognize this as

\frac{d}{d x} (9^{x} - 5^{x}) ∣_{x = 0}

That is:

lim_{x \to 0} \frac{9^{x} - 5^{x}}{x} = lim_{x \to 0} (\frac{e^{x \log {9}} - e^{0 \cdot \log {9}}}{x} - \frac{e^{x \log {5}} - e^{0 \cdot \log {5}}}{x})

= \frac{d}{d x} e^{x \log {9}} {|_{{x = 0}} - \frac{d}{d x} e^{x \log {5}} |}_{x = 0}

by definition of the derivative at zero.

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