Question

The value of $\underset{x\to 0}{lim}\frac{x^2{e}^x}{cosx-1}$ using taylor series is?

• A. $2$
• B. $-3$
• C. $-2$
• D. $1$

Answer 1

Answer: (C)

$-2$

${lim}_{x\to 0}\frac{x^2{e}^x}{cosx-1}=?$
We need to find the limit value using Taylor series.
The Taylor series of $x^2$ is $x^2$
The Taylor series of $e^x$ is $1+x+\frac{1}{2}{x}^2+...$
The Taylor series of $cosx$ is $1-\frac{1}{2}{x}^2+\frac{1}{24}{x}^4-...$
${lim}_{x\to 0}\frac{x^2{e}^x}{cosx-1}\approx {lim}_{x\to 0}\frac{x^2(1+x+\frac{1}{2}{x}^2+...)}{1-\frac{1}{2}{x}^2+\frac{1}{24}{x}^4-...-1}$
$={lim}_{x\to 0}\frac{x^2(1+x+\frac{1}{2}{x}^2)}{1-\frac{1}{2}{x}^2+\frac{1}{24}{x}^4-1}$
$={lim}_{x\to 0}\frac{x^2(1+x+\frac{1}{2}{x}^2)}{-\frac{1}{2}{x}^2+\frac{1}{24}{x}^4}$
$={lim}_{x\to 0}\frac{x^2(1+x+\frac{1}{2}{x}^2)}{x^2(-\frac{1}{2}+\frac{1}{24}{x}^2)}$
$={lim}_{x\to 0}\frac{1+x+\frac{1}{2}{x}^2}{-\frac{1}{2}+\frac{1}{24}{x}^2}$
$={lim}_{x\to 0}\frac{1}{-\frac{1}{2}}$
${lim}_{x\to 0}\frac{x^2{e}^x}{cosx-1}=-2$