Question

Is there a way to solve this without l'hopital's rule?

Answer 1

The conclusion is that the expression $\lim_{x \to \infty} \frac{6 (2^x) - 1}{5x^2+ 10}$ cannot be solved  without the l'hopital's rule,

How to solve the limit expression?

The limit expression is given as:

$\lim_{x \to \infty} \frac{6 (2^x) - 1}{5x^2+ 10}$

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.

$\lim_{x \to \infty} \frac{6 (2^{\infty}) - 1}{5(\infty)^2+ 10} = \frac{\infty}{\infty}$

So, the best way is to apply the l'hopital's rule.

This is done as follows:

$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)}$

When the numerator and the denominator are differentiated, we have:

$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \frac{3\ln(2)\cdot 2^{x + 1}}{10x}$

Further, differentiate

$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \frac{3\ln^2(2)\cdot 2^{x + 1}}{10}$

Now, we can substitute $\infty$ for x

$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \frac{3\ln^2(2)\cdot 2^{\infty + 1}}{10}$

Evaluate the numerator

$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \frac{\infty}{10}$

Evaluate the quotient

$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \infty$

Hence, the conclusion is that the expression $\lim_{x \to \infty} \frac{6 (2^x) - 1}{5x^2+ 10}$ 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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