Question

For a certain value of $c,\underset{x\to -\infty }{\lim }\left[{\left(x^5+7x^4+2\right)}^c-x\right]=\lambda$, is finite and non-zero. Then, the value of $3c+\lambda$ is

Answer 1

Answer: 2

$\underset{x\to -\infty }{\lim }\left(x^{5c}{\left(1+\frac{7}{x}+\frac{2}{x^5}\right)}^c-x\right)$
$=\underset{x\to -\infty }{\lim }x\left(x^{5c-1}{\left(1+\frac{7}{x}+\frac{2}{x^5}\right)}^c-1\right)$
This will be of the form $\infty \times 0$ only, if
$5c-1=0$
$\Rightarrow c=\frac{1}{5}$ substituting $c=\frac{1}{5},$
$\lambda$ becomes
Thus, $\lambda =\underset{x\to -\infty }{\lim }\left[(1+x{)}^{1/5}-1\right]$
where $x=\frac{7}{x}+\frac{2}{x^5}$
$=\underset{x\to -\infty }{\lim }x\left[1+\frac{x}{5}+\dots .-1\right]$
$=\underset{x\to -\infty }{\lim }x\left(\frac{7}{x}+\frac{2}{x^5}\right)\cdot \frac{1}{5}=\frac{7}{5}$
Hence, $c=\frac{1}{5}$ and $\lambda =\frac{7}{5}$
$\Rightarrow 3c+\lambda =2$