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d)
lim_(x rarr0)(e^(x)-e^(-x)-2x)/(x-sin x)

d) $\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}-2 x}{x-\sin x}$

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Simplify rational expressions

Full solution

Q. d)

\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}-2 x}{x-\sin x}

Plug in $x = 0$ : First, let's try to plug in $x = 0$ directly and see what happens. $\newline$ $\lim_{x \to 0}\frac{e^{x}-e^{-x}-2x}{x-\sin x} = \frac{e^{0}-e^{0}-2\cdot 0}{0-\sin 0} = \frac{1-1-0}{0-0} = \frac{0}{0}$ $\newline$ We get an indeterminate form, so we need to use L'Hôpital's Rule.
Apply L'Hôpital's Rule: L'Hôpital's Rule says that if we have an indeterminate form of $0/0$ or $\infty/\infty$ , we can take the derivative of the numerator and the denominator separately and then take the limit. $\newline$ So, let's find the derivative of the numerator: $\frac{d}{dx}(e^{x}-e^{-x}-2x) = e^{x} + e^{-x} - 2$ .
Find derivative of numerator: Now, let's find the derivative of the denominator: $\frac{d}{dx}(x-\sin x) = 1 - \cos x$ .
Find derivative of denominator: Now we take the limit of the derivatives: $\lim_{x \to 0}\frac{e^{x} + e^{-x} - 2}{1 - \cos x}$ . Let's plug in $x = 0$ again: $\frac{e^{0} + e^{0} - 2}{1 - \cos 0} = \frac{1 + 1 - 2}{1 - 1} = \frac{0}{0}$ . Oops, we got another indeterminate form, so we need to apply L'Hôpital's Rule again.

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