Question

Lim x -> infinity (4x+1/4x)^2x​

Answer 1

Применим второй замечательный предел

$\displaystyle \lim_{x\to \infty}\left(\frac{4x+1}{4x}\right)^\big{2x}=\lim_{x\to \infty}\left(1+\frac{1}{4x}\right)^\big{4x\cdot 0.5}=e^{0.5}=\sqrt{e}$

Answer 2

To evaluate the limit as x approaches infinity of the expression (4x + 1)/(4x)^(2x), we can use some algebraic manipulations and properties of limits. Let's simplify the expression first:

(4x + 1)/(4x)^(2x)

= [(4x + 1)/(4x)]^(2x) [Using the property (a/b)^c = a^c / b^c]

= [(1 + 4x)/(4x)]^(2x) [Dividing both numerator and denominator by 4x]

Now, let's focus on the limit:

lim(x->∞) [(1 + 4x)/(4x)]^(2x)

This expression is in the form of (1^∞), which is an indeterminate form. To solve it, we can take the natural logarithm (ln) of both sides:

ln [lim(x->∞) [(1 + 4x)/(4x)]^(2x)]

= ln [lim(x->∞) (1 + 4x)^(2x) / (4x)^(2x)]

= 2x ln (1 + 4x) - 2x ln (4x)

Now, we can simplify further. As x approaches infinity, the dominant term in the logarithms will be the one with the highest exponent. In this case, it is 4x.

ln [lim(x->∞) [(1 + 4x)/(4x)]^(2x)]

≈ 2x ln (4x) - 2x ln (4x) [Taking the limit as x approaches infinity]

≈ 0

Finally, taking the exponential of both sides, we have:

lim(x->∞) [(1 + 4x)/(4x)]^(2x) ≈ e^0

lim(x->∞) [(1 + 4x)/(4x)]^(2x) ≈ 1

Therefore, as x approaches infinity, the given expression converges to 1.