Answer:
C is not 0.
[tex]\boxed{C=\dfrac{625}{24}}[/tex]
[tex]\lim_{x \to \infty} \dfrac{e^{5x}}{x^5} = \lim_{x \to \infty} \dfrac{625}{24}e^{5x} = \infty[/tex]
Step-by-step explanation:
According to the L'Hospital's Rule, when we have an indetermination of the form
[tex]\dfrac{0}{0} \text{ or } \dfrac{\pm\infty}{\pm\infty}[/tex]
So,
[tex]\[ \lim_{x\to\infty} \dfrac{f(x)}{g(x)} = \lim_{x\to\infty} \dfrac{f'(x)}{g'(x)} \][/tex]
In this question,
[tex]\dfrac{d}{dx}e^{5x}=5e^{5x}[/tex]
[tex]\dfrac{d}{dx}x^{5}=5x^{4}[/tex]
So,
[tex]\lim_{x \to \infty} \dfrac{e^{5x}}{x^5} = \lim_{x \to \infty} \dfrac{e^{5x}}{x^4}[/tex]
But it still in the indeterminate form, so we repeat the rule again.
[tex]\dfrac{d}{dx}e^{5x}=5e^{5x}[/tex]
[tex]\dfrac{d}{dx}x^{4}=4x^{3}[/tex]
So,
[tex]\lim_{x \to \infty} \dfrac{e^{5x}}{x^5} = \lim_{x \to \infty} \dfrac{5e^{5x}}{4x^3}[/tex]
But it still in the indeterminate form, so we repeat the rule again.
[tex]\dfrac{d}{dx}5e^{5x}=25e^{5x}[/tex]
[tex]\dfrac{d}{dx}4x^{3}=12x^{2}[/tex]
So,
[tex]\lim_{x \to \infty} \dfrac{e^{5x}}{x^5} = \lim_{x \to \infty} \dfrac{25e^{5x}}{12x^2}[/tex]
But it still in the indeterminate form, so we repeat the rule again.
[tex]\dfrac{d}{dx}25e^{5x}=125e^{5x}[/tex]
[tex]\dfrac{d}{dx}12x^{2}=24x[/tex]
So,
[tex]\lim_{x \to \infty} \dfrac{e^{5x}}{x^5} = \lim_{x \to \infty} \dfrac{125e^{5x}}{24x}[/tex]
But it still in the indeterminate form, so we repeat the rule again. I will be the last time.
[tex]\dfrac{d}{dx}125e^{5x}=625e^{5x}[/tex]
[tex]\dfrac{d}{dx}24x=24[/tex]
So,
[tex]\lim_{x \to \infty} \dfrac{e^{5x}}{x^5} = \lim_{x \to \infty} \dfrac{625e^{5x}}{24}[/tex]
Now we can evaluate the limit.
So, [tex]\boxed{C=\dfrac{625}{24}}[/tex]
[tex]\lim_{x \to \infty} \dfrac{e^{5x}}{x^5} = \lim_{x \to \infty} \dfrac{625}{24}e^{5x} = \infty[/tex]
