Question

For a certain value of $c$, ${lim}_{x\to \infty }\left[\right(x^5+7x^4+2{)}^c-x]$ is finite & non zero. The value of $c$ and the value of the limit is

• A. $\frac{1}{5},\frac{7}{5}$
• B. 0, 1
• C. $1,\frac{7}{5}$
• D. none

Answer 1

Answer: (A)

$\frac{1}{5},\frac{7}{5}$

$\underset{x\to \infty }{lim}[{(x^5+{7x}^4+2)}^c-x]=\underset{x\to \infty }{lim}[x^{5c}{(1+\frac{7}{x}+\frac{2}{x^5})}^c-x]$

For limit to exits $c=\frac{1}{5}$

Now to find limit put $c=\frac{1}{5}$

$\Rightarrow \underset{x\to \infty }{lim}[x{(1+\frac{7}{x}+\frac{2}{x^5})}^{\frac{1}{5}}-x]=\frac{7}{5}$ ( by expansion)