The conclusion is that the expression [tex]\lim_{x \to \infty} \frac{6 (2^x) - 1}{5x^2+ 10}[/tex] cannot be solved without the l'hopital's rule,
How to solve the limit expression?
The limit expression is given as:
[tex]\lim_{x \to \infty} \frac{6 (2^x) - 1}{5x^2+ 10}[/tex]
The limit of the expression cannot be solved without l'hopital's rule, because a direct substitution of ∞ for x would result in the following expression ∞/∞
i.e.
[tex]\lim_{x \to \infty} \frac{6 (2^{\infty}) - 1}{5(\infty)^2+ 10} = \frac{\infty}{\infty}[/tex]
So, the best way is to apply the l'hopital's rule.
This is done as follows:
[tex]\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)}[/tex]
When the numerator and the denominator are differentiated, we have:
[tex]\lim_{x \to \infty} \frac{f(x)}{g(x)} = \frac{3\ln(2)\cdot 2^{x + 1}}{10x}[/tex]
Further, differentiate
[tex]\lim_{x \to \infty} \frac{f(x)}{g(x)} = \frac{3\ln^2(2)\cdot 2^{x + 1}}{10}[/tex]
Now, we can substitute [tex]\infty[/tex] for x
[tex]\lim_{x \to \infty} \frac{f(x)}{g(x)} = \frac{3\ln^2(2)\cdot 2^{\infty + 1}}{10}[/tex]
Evaluate the numerator
[tex]\lim_{x \to \infty} \frac{f(x)}{g(x)} = \frac{\infty}{10}[/tex]
Evaluate the quotient
[tex]\lim_{x \to \infty} \frac{f(x)}{g(x)} = \infty[/tex]
Hence, the conclusion is that the expression [tex]\lim_{x \to \infty} \frac{6 (2^x) - 1}{5x^2+ 10}[/tex] cannot be solved without the l'hopital's rule
Read more about l'hopital's rule at:
https://brainly.com/question/2095652
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