For a certain value of c, mathop {Lim}limits_ | vedclass

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(A) Consider the limit $L = \mathop {Lim}\limits_{x 	o - \infty } [(x^5 + 7x^4 + 2)^c - x]$.For the limit to be finite and non-zero, we factor out $x

Vedclass · Accepted Answer

$1/5, 7/5$

Vedclass · Answer

(A) Consider the limit $L = \mathop {Lim}\limits_{x 	o - \infty } [(x^5 + 7x^4 + 2)^c - x]$.For the limit to be finite and non-zero, we factor out $x^{5c}$:$L = \mathop {Lim}\limits_{x 	o - \infty } [x^{5c}(1 + \frac{7}{x} + \frac{2}{x^5})^c - x]$.For this to be of the form $\infty 	imes 0$, we must have $5c = 1$, so $c = 1/5$.Substituting $c = 1/5$:$L = \mathop {Lim}\limits_{x 	o - \infty } x[(1 + \frac{7}{x} + \frac{2}{x^5})^{1/5} - 1]$.Using the binomial expansion $(1+u)^n \approx 1 + nu$ for small $u$:$L = \mathop {Lim}\limits_{x 	o - \infty } x[1 + \frac{1}{5}(\frac{7}{x} + \frac{2}{x^5}) - 1]$.$L = \mathop {Lim}\limits_{x 	o - \infty } x[\frac{7}{5x} + \frac{2}{5x^5}] = \frac{7}{5}$.Thus, $c = 1/5$ and the limit is $7/5$.

For a certain value of $c$, $\mathop {Lim}\limits_{x \to - \infty } [(x^5 + 7x^4 + 2)^c - x]$ is finite and non-zero. The value of $c$ and the value of the limit are:

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