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For a certain value of $'c',\lim_{x\rightarrow \infty}\left[\left(x^5+7x^4+2\right)^c-x\right]$ is a finite and non-zero,

Then value of limit is

$\bf{My\; Try::}$ Let $\displaystyle x=\frac{1}{y}\;,$ Then $\displaystyle \lim_{y\rightarrow 0}\left[\left(\frac{2y^5+7y+1}{y^5}\right)^c-\frac{1}{y}\right]=\lim_{y\rightarrow 0}\frac{y\cdot \left(2y^5+7y+1\right)^c-y^{5c}}{y^{5c}\cdot y}$

Now how can I solve after that,Help me, Thanks

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If $$\lim_{x\to\infty}\bigl((x^5+7x^4+2)^c-x\bigr)$$ is finite, then $$\lim_{x\to\infty}\frac{(x^5+7x^4+2)^c-x}{x}=0$$ and therefore $$\lim_{x\to\infty}\frac{(x^5+7x^4+2)^c}{x}=1\ .$$ Hence $$\lim_{x\to\infty}\frac{x^5+7x^4+2}{x^{1/c}}=1\ .$$ Now if $\frac1c>5$ then the limit is zero; if $\frac1c<5$ then the limit is infinite; so we must have $c=\frac15$.

To evaluate the limit, write $u=(x^5+7x^4+2)^{1/5}$. Then we have $$7x^4+2=u^5-x^5=(u-x)(u^4+u^3x+u^2x^2+ux^3+x^4)$$ and so $$u-x=\frac{7x^4+2}{u^4+u^3x+u^2x^2+ux^3+x^4}\ .\tag{$*$}$$ Now all the terms in the denominator are much the same as $x\to\infty$: I'll do one and leave you to think about the rest. We have $$u^3x=x^4\Bigl(1+\frac7x+\frac{2}{x^5}\Bigr)^{3/5}\ ,$$ and the term in brackets tends to $1$ as $x\to\infty$. So we can cancel $x^4$ from top and bottom of $(*)$ to get $$\lim_{x\to\infty}(u-x)=\frac75\ .$$

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$\begin{array}\\ (x^5+7x^4+2)^c &=(x^5(1+7/x+2/x^5))^c\\ &=x^{5c}(1+7/x+2/x^5)^c\\ &\approx x^{5c}(1+c(7/x+2/x^5) + O(1/x^2))\\ &= x^{5c}(1+7c/x + O(1/x^2))\\ &= x^{5c}+7cx^{5c-1} + O(x^{5c-2})\\ \end{array} $

Therefore $(x^5+7x^4+2)^c-x =x^{5c}+7cx^{5c-1} + O(x^{5c-2})-x $

For this to be finite for large $x$, the $x^{5c}$ and $x$ terms must cancel, so $5c=1$ or $c=\frac15$.

The remaining term is $7cx^{5c-1} =\frac75 $ since $5c-1 = 0 $.

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