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

A)$-1$;

B)$0$;

C)$1$;

D)None of these

Ioriannis4v

Beginner2022-12-26Added 11 answers

The correct answer is C $1$

Explanation for the correct answer:

Applying the limits and reducing the equation to its simplest form:

$\Rightarrow \mathrm{lim}x\to 0\left(\frac{{e}^{x}-{e}^{\mathrm{sin}x}}{x-\mathrm{sin}x}\right)\Rightarrow \mathrm{lim}x\to 0\left(\frac{{e}^{x}-{e}^{\mathrm{sin}x}\mathrm{cos}x}{1-\mathrm{cos}x}\right)\left(\right)\Rightarrow \mathrm{lim}x\to 0\left(\frac{{e}^{x}-\left({e}^{\mathrm{sin}x}\left(-\mathrm{sin}x\right)+{e}^{\mathrm{sin}x}\mathrm{cos}x\xb7\mathrm{cos}x\right)}{0-\left(-\mathrm{sin}x\right)}\right)\left(\right)\Rightarrow \mathrm{lim}x\to 0\left(\frac{{e}^{x}+{e}^{\mathrm{sin}x}.\mathrm{sin}x-{\mathrm{cos}}^{2}x\xb7x{e}^{\mathrm{sin}x}}{\mathrm{sin}x}\right)\Rightarrow \mathrm{lim}x\to 0\left(\frac{{e}^{x}+{e}^{\mathrm{sin}x}\mathrm{cos}x+\mathrm{sin}x\xb7{e}^{\mathrm{sin}x}-{\mathrm{cos}}^{2}x\xb7{e}^{\mathrm{sin}x}{\mathrm{cos}}^{x-2\mathrm{cos}x\left(-\mathrm{sin}x\right){e}^{\mathrm{sin}x}}}{\mathrm{cos}x}\right)\left(\right)$

Applying the limits

$\Rightarrow \frac{{e}^{0}+{e}^{0}\times 1+0\xb7{e}^{0}\xb71-{1}^{2}\xb7{e}^{0}\xb71-2\xb71\left(-0\right){e}^{0}}{1}\Rightarrow 1+1-1\Rightarrow 1$

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.

Explanation for the correct answer:

Applying the limits and reducing the equation to its simplest form:

$\Rightarrow \mathrm{lim}x\to 0\left(\frac{{e}^{x}-{e}^{\mathrm{sin}x}}{x-\mathrm{sin}x}\right)\Rightarrow \mathrm{lim}x\to 0\left(\frac{{e}^{x}-{e}^{\mathrm{sin}x}\mathrm{cos}x}{1-\mathrm{cos}x}\right)\left(\right)\Rightarrow \mathrm{lim}x\to 0\left(\frac{{e}^{x}-\left({e}^{\mathrm{sin}x}\left(-\mathrm{sin}x\right)+{e}^{\mathrm{sin}x}\mathrm{cos}x\xb7\mathrm{cos}x\right)}{0-\left(-\mathrm{sin}x\right)}\right)\left(\right)\Rightarrow \mathrm{lim}x\to 0\left(\frac{{e}^{x}+{e}^{\mathrm{sin}x}.\mathrm{sin}x-{\mathrm{cos}}^{2}x\xb7x{e}^{\mathrm{sin}x}}{\mathrm{sin}x}\right)\Rightarrow \mathrm{lim}x\to 0\left(\frac{{e}^{x}+{e}^{\mathrm{sin}x}\mathrm{cos}x+\mathrm{sin}x\xb7{e}^{\mathrm{sin}x}-{\mathrm{cos}}^{2}x\xb7{e}^{\mathrm{sin}x}{\mathrm{cos}}^{x-2\mathrm{cos}x\left(-\mathrm{sin}x\right){e}^{\mathrm{sin}x}}}{\mathrm{cos}x}\right)\left(\right)$

Applying the limits

$\Rightarrow \frac{{e}^{0}+{e}^{0}\times 1+0\xb7{e}^{0}\xb71-{1}^{2}\xb7{e}^{0}\xb71-2\xb71\left(-0\right){e}^{0}}{1}\Rightarrow 1+1-1\Rightarrow 1$

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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