Question

If $\underset{x\to 0}{\lim }\frac{\alpha x(1+x+{\displaystyle \frac{x^2}{2!}}+{\displaystyle \frac{x^3}{3!}}+...)-\beta (x-{\displaystyle \frac{x^2}{2}}+{\displaystyle \frac{x^3}{3}}+...)+\gamma x^2(1-x+{\displaystyle \frac{x^{2^{}}}{2!}}-{\displaystyle \frac{x^3}{3!}}+...)}{x^3}=10$ .

Find the value of $\alpha +\beta +\gamma$

Answer 1

Finding the value of $\mathit{\alpha }\mathbf{+}\mathit{\beta }\mathbf{+}\mathit{\gamma }$ :

Given that $\underset{x\to 0}{\lim }\frac{\alpha x(1+x+{\displaystyle \frac{x^2}{2!}}+{\displaystyle \frac{x^3}{3!}}+...)-\beta (x-{\displaystyle \frac{x^2}{2}}+{\displaystyle \frac{x^3}{3}}+...)+\gamma x^2(1-x+{\displaystyle \frac{x^{2^{}}}{2!}}-{\displaystyle \frac{x^3}{3!}}+...)}{x^3}=10$

For limit to exist

$\alpha –\beta =0,\alpha +\frac{\beta }{2}+\gamma =0$, $\frac{\alpha }{2}–\frac{\beta }{3}–\gamma =10$…(i)

$\Rightarrow$ $\alpha =\beta ,\gamma =-3\frac{\alpha }{2}$

Put the value of $\gamma$ and $\beta$ in equation (i)

$\frac{\alpha }{2}-\frac{\alpha }{3}+\frac{3\alpha }{2}=10$

$\Rightarrow$ $\frac{\alpha }{6}+\frac{3\alpha }{2}=10$

$\Rightarrow$ $\frac{\alpha +9\alpha }{6}=10$

$\Rightarrow$ $\alpha =6$

$\Rightarrow$$\alpha =6,\beta =6,\gamma =-9$

$\begin{array}{rcl}\therefore \alpha +\beta +\gamma & =& 6+6-9\\ & =& 3\end{array}$

Hence, The value of $\alpha +\beta +\gamma$ is $\mathbf{3}$