How do you find the limit of #((16x)/(16x+3))^(4x)# as x approaches infinity?

So

Consequently,

To find the limit of ((16x)/(16x+3))^(4x) as x approaches infinity, we can use the properties of limits and exponential functions.

First, we rewrite the expression as e^(ln(((16x)/(16x+3))^(4x))).

Next, we simplify the exponent by using the properties of logarithms: ln(((16x)/(16x+3))^(4x)) = 4x * ln((16x)/(16x+3)).

Now, we can evaluate the limit of the exponent as x approaches infinity.

Using the limit properties, we can rewrite the expression as e^(lim(x→∞) [4x * ln((16x)/(16x+3))]).

Next, we evaluate the limit of the logarithm term: lim(x→∞) ln((16x)/(16x+3)).

By applying the properties of logarithms, we can simplify this to lim(x→∞) ln(16x) - ln(16x+3).

As x approaches infinity, ln(16x) and ln(16x+3) both tend to infinity.

Therefore, the limit of ln(16x) - ln(16x+3) as x approaches infinity is infinity.

Finally, we have e^(lim(x→∞) [4x * ln((16x)/(16x+3))]) = e^(∞) = infinity.

Hence, the limit of ((16x)/(16x+3))^(4x) as x approaches infinity is infinity.

To find the limit of ( \left(\frac{16x}{16x+3}\right)^{4x} ) as ( x ) approaches infinity, we can rewrite the expression using properties of logarithms. Taking the natural logarithm of both sides, we have:

[ \lim_{x \to \infty} \left(\frac{16x}{16x+3}\right)^{4x} = \lim_{x \to \infty} \ln \left(\left(\frac{16x}{16x+3}\right)^{4x}\right) ]

Using the properties of logarithms, we can bring the exponent down:

[ = \lim_{x \to \infty} 4x \ln \left(\frac{16x}{16x+3}\right) ]

Next, we can use L'Hôpital's Rule, which states that if we have an indeterminate form of ( \frac{0}{0} ) or ( \frac{\infty}{\infty} ), we can take the derivative of the numerator and the derivative of the denominator and then find the limit again.

So, we differentiate both the numerator and the denominator:

[ \lim_{x \to \infty} 4x \ln \left(\frac{16x}{16x+3}\right) = \lim_{x \to \infty} \frac{4 \ln \left(\frac{16x}{16x+3}\right)}{\frac{1}{x}} ]

Applying L'Hôpital's Rule:

[ = \lim_{x \to \infty} \frac{4 \left(\frac{d}{dx}\ln \left(\frac{16x}{16x+3}\right)\right)}{\left(\frac{d}{dx}\frac{1}{x}\right)} ]

[ = \lim_{x \to \infty} \frac{4 \left(\frac{1}{\frac{16x}{16x+3}} \cdot \frac{d}{dx}\left(\frac{16x}{16x+3}\right)\right)}{-\frac{1}{x^2}} ]

[ = \lim_{x \to \infty} \frac{4 \left(\frac{16(16x+3) - 16x \cdot 16}{(16x+3)^2}\right)}{-\frac{1}{x^2}} ]

[ = \lim_{x \to \infty} \frac{4 \cdot \frac{768}{(16x+3)^2}}{-\frac{1}{x^2}} ]

[ = \lim_{x \to \infty} \frac{-3072x^2}{(16x+3)^2} ]

Now, as ( x ) approaches infinity, the denominator grows faster than the numerator, so the limit approaches zero:

[ \lim_{x \to \infty} \frac{-3072x^2}{(16x+3)^2} = 0 ]

Therefore, the limit of ( \left(\frac{16x}{16x+3}\right)^{4x} ) as ( x ) approaches infinity is 0.