Linda Norman

2022-12-25

Evaluate:$\mathrm{lim}x\to 0\frac{{e}^{x}-{e}^{\mathrm{sin}x}}{x-\mathrm{sin}x}$
A)$-1$;
B)$0$;
C)$1$;
D)None of these

Ioriannis4v

The correct answer is C $1$
Applying the limits and reducing the equation to its simplest form:
$⇒\mathrm{lim}x\to 0\left(\frac{{e}^{x}-{e}^{\mathrm{sin}x}}{x-\mathrm{sin}x}\right)⇒\mathrm{lim}x\to 0\left(\frac{{e}^{x}-{e}^{\mathrm{sin}x}\mathrm{cos}x}{1-\mathrm{cos}x}\right)\left(Differentiating\right)⇒\mathrm{lim}x\to 0\left(\frac{{e}^{x}-\left({e}^{\mathrm{sin}x}\left(-\mathrm{sin}x\right)+{e}^{\mathrm{sin}x}\mathrm{cos}x·\mathrm{cos}x\right)}{0-\left(-\mathrm{sin}x\right)}\right)\left(Differentiating\right)⇒\mathrm{lim}x\to 0\left(\frac{{e}^{x}+{e}^{\mathrm{sin}x}.\mathrm{sin}x-{\mathrm{cos}}^{2}x·x{e}^{\mathrm{sin}x}}{\mathrm{sin}x}\right)⇒\mathrm{lim}x\to 0\left(\frac{{e}^{x}+{e}^{\mathrm{sin}x}\mathrm{cos}x+\mathrm{sin}x·{e}^{\mathrm{sin}x}-{\mathrm{cos}}^{2}x·{e}^{\mathrm{sin}x}{\mathrm{cos}}^{x-2\mathrm{cos}x\left(-\mathrm{sin}x\right){e}^{\mathrm{sin}x}}}{\mathrm{cos}x}\right)\left(Differentiating\right)$
Applying the limits
$⇒\frac{{e}^{0}+{e}^{0}×1+0·{e}^{0}·1-{1}^{2}·{e}^{0}·1-2·1\left(-0\right){e}^{0}}{1}⇒1+1-1⇒1$
Thus, $\mathrm{lim}x\to 0\frac{{e}^{x}-{e}^{\mathrm{sin}x}}{x-\mathrm{sin}x}=1$
Option (C) is, thus, the right response.

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