Find a constant so the given function has a finite, non-zero limit

Find the value for the constant $c$ so that

$\lim_{x \to +\infty} \left( (x^5 + 7x^4 + 2)^c - x \right)$

is finite and is not equal to zero. Compute the value of the limit.

$\lim_{x \to 0} (1+x)^c = 1 + cx + o(x).$

To apply this result, we first make the substitution $t = \frac{1}{x}$ , then $t \to 0$ as $x \to +\infty$ and we have

$\begin{align*} \lim_{x \to +\infty} \left( (x^5 + 7x^4 + 2)^c - x \right) &= \lim_{t \to 0} \left(\left( \frac{1}{t^5} + \frac{7}{t^4} + 2 \right)^c - \frac{1}{t} \right) \\[9pt] &= \lim_{t \to 0} \left( \frac{1}{t^{5c}} \left( 1 + (7t + 2t^5) \right)^c - \frac{1}{t} \right) \\[9pt] &= \lim_{t \to 0} \left( \frac{(1 + (7t+2t^5))^c - t^{5c-1}}{t^{5c}} \right). \end{align*}$

Now, since $(7t + 2t^5) \to 0$ as $t \to 0$ we can apply the result of the previous exercise we stated above which tells us

$\lim_{t \to 0} (1+(7t+2t^5))^c = 1 + c(7t+2t^5) + o(7t+2t^5) = 1 + c(7t+2t^5) + o(t).$

Therefore,

$\lim_{t \to 0} \frac{ (1+(7t+2t^5))^c - t^{5c-1} }{t^{5c}} &= \lim_{t \to 0} \frac{1 + c (7t+2t^5) + o(t) - t^{5c-1}}{t^{5c}}.$

Now, for this limit to be finite and non-zero we need the term in the numerator to have no constant terms (otherwise the $t^{5c}$ in the denominator will make the limit infinite). So, we need $1 - t^{5c-1} = 0$ which means $5c - 1=0$ or $c = \frac{1}{5}$ . Substituting this value of $c$ we then evaluate the limit

$\begin{align*} \lim_{t \to 0} \frac{ 1 + \frac{1}{5} (7t+2t^5) + o(t) - 1}{t} &= \lim_{t \to 0} \left( \frac{7}{5} + \frac{2}{5}t^4 + \frac{o(t)}{t} \right) \\[9pt] &= \frac{7}{5}. \end{align*}$

Stumbling Robot

Find a constant so the given function has a finite, non-zero limit

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