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  • Evaluating     <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{e^x \sin x-x(x+1)}{x^3}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightarr

Evaluating     \mathrm{\lim _{x \rightarrow 0} \frac{e^x \sin x-x(x+1)}{x^3}}

Option: 1

0


Option: 2

1/3


Option: 3

1


Option: 4

-1


Answers (1)

best_answer

Since the denominator is \mathrm{x^3}, you have to at least expand the exponential and the sine function up to order three. Consider this:
\mathrm{ e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+o\left(x^3\right) }
\mathrm{ \sin (x)=x-\frac{x^3}{6}+o\left(x^3\right) }
hence
\mathrm{ \sin (x)=x-\frac{x^3}{6}+o\left(x^3\right) e^x}

\mathrm{ \sin (x)=\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+o\left(x^3\right)\right)\left(x-\frac{x^3}{6}+o\left(x^3\right)\right)=x+x^2-\frac{x^3}{6}+\frac{x^3}{2}+o\left(x^3\right) }

That is
\mathrm{ e^x \sin (x)=x+x^2+\frac{x^3}{3}+o\left(x^3\right)}
Where we have trashed terms of order higher than  \mathrm{x^3.}
Then we have
$$
\mathrm{\lim _{x \rightarrow 0} \frac{x+x^2+\frac{x^3}{3}+o\left(x^3\right)-x^2-x}{x^3}=\lim _{x \rightarrow 0} \frac{\frac{x^3}{3}+o\left(x^3\right)}{x^3}}

\mathrm{=\lim _{x \rightarrow 0} \frac{1}{3}+\underbrace{\lim _{x \rightarrow 0} \frac{o\left(x^3\right)}{x^3}}_{\rightarrow 0}=\frac{1}{3}}.

Posted by

vishal kumar

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