Question

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

Answer 1

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