Limit and derivatives

Calculus Level 2

$\large \lim_{x \to 8} \frac d{dx} \left(\frac{e^{2x} - 1}{e^x +1} \right) = ?$

Submit your answer to three decimal places.

The answer is 2980.957.

1 solution

Anirudh Sreekumar
Dec 13, 2017

$\begin{aligned} \lim_{x \to 8} \frac d{dx} \left(\frac{e^{2x} - 1}{e^x +1} \right)&=\lim_{x \to 8} \frac d{dx} \left(\frac{(e^{x})^2 - 1^2}{e^x +1} \right)\\ &=\lim_{x \to 8} \frac d{dx} (e^x-1) \\\\ &=\lim_{x \to 8} e^x\\\\ &=e^8\\\\ &\approx\color{#EC7300}\boxed{\color{#333333}2980.957}\end{aligned}$

Limit and derivatives

The answer is 2980.957.

1 solution

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