$\lim_{x\to \infty} (x^4 +5x^3+3)^c-x=k$. Find $k$ and $c$. 
$\lim_{x\to \infty} (x^4 +5x^3+3)^c-x=k$. Find $k$ and $c$.

key points: This is supposed to be solved with knowledge of just L'hospital's rule.
*The value of limit i.e, k is finite and non zero
I have tried manipulation using logarithm which didn't turn out to be of good use. I think multiplying by conjugate should work but that just  gives complex derivatives  and doesn't seem to lead to any answer.
Am I missing some basic point or is there a specific procedure for this problem.
 A: $$L=\lim_{x\to \infty} (x^4 +5x^3+3)^c-x=k$$.
Here $c$ needs to be $1/4$ so that asymptotically the first part in above go as $x$, hence
$$L=\lim_{x\to \infty} (x^4 +5x^3+3)^{1/4}-x=k$$.
$$L=\lim_{x\to \infty} x (1 +5/x+3/x^4)^{1/4}-x=k$$.
Apply binomial expansion
$(1+z)^{p}=1+pz+O(z^2)+...$ id $|z|<<1$
$$L=\lim_{x \to \infty} (x+5/4+3/(4x^3)+...)-x=5/4=k$$
A: First you may find $c$ by considering that
$$0=\lim_{x\to \infty}\frac kx =\lim_{x\to \infty}\frac{(x^4+5x^3+3)^c - x}{x}= \lim_{x\to \infty}\left(x^{4c-1}(1+\frac 5x+\frac 3{x^4})^c - 1\right)$$
Hence, $\lim_{x\to \infty }x^{4c-1}=1\Leftrightarrow c=\frac 14$.
Now, you can use L'Hospital to find
\begin{eqnarray*}\sqrt[4]{x^4+5x^3+3} - x
& = & \frac{\sqrt[4]{1+\frac 5x + \frac 3{x^4}}-1}{\frac 1x} \\
& \stackrel{t=\frac 1x}{=} &  \frac{\sqrt[4]{1+5t + 3t^4}-1}{t} \\
& \stackrel{L'Hosp.,\; t\to 0^+}{\sim} & \frac{5+12t^3}{4\sqrt[4]{(1+5t + 3t^4)^3}}\\
& \stackrel{t\to 0^+}{\longrightarrow} & \frac 54
\end{eqnarray*}
A: Hints: If $c <\frac 1  4$ then the limit on the left is $-\infty$.
If $c>\frac 1 4 $ write LHS as $\lim x^{4c} [(1+\frac 5  x+\frac 3 {x^{4}})^{c} -x^{1-4c}]$ to see that the limit is $+\infty$.
Finally, let $c=\frac 1  4$. Then LHS is $\lim x[(1+5x^{-1}+3x^{-4})^{1/4}-1]$. This limit is $\frac 5  4$ so we must have $c=\frac 1 4$ and $k=\frac  5 4 $.
For the case $c=\frac 1  4$ bring $x$ to he denominator and apply L'Hopital's Rule.
A: Set $1/x=h$ to find
$$S=\lim_{h\to0^+}\left(\dfrac{(1+5h+3h^4)^c}{h^{4c}}-\dfrac1h\right)$$
If $4c=1+d, d>0$
$$S=\lim_{h\to0^+}\left(\dfrac{(1+5h+3h^4)^{\frac{d+1}4}}{h^{1+d}}-\dfrac1h\right)=\lim_{h\to0^+}\dfrac{(1+5h+3h^4)^{\frac{d+1}4}-h^d}{h^{1+d}}=?$$
What if $4c=1-d, d>0?$
So, we need to have $4c=1$
$$S=\lim_{h\to0^+}\left(\dfrac{(1+h(5+3h^3))^{1/4}-1}h\right)$$
Use Binomial series  to find $$S=\dfrac54$$
