# How do you evaluate #((2x-1)/(2x+5))^(2x+3)# as x approaches infinity?

but

then

and

From any Calculus textbook we get

so

By signing up, you agree to our Terms of Service and Privacy Policy

To evaluate ((2x-1)/(2x+5))^(2x+3) as x approaches infinity, we can use the concept of limits.

First, we rewrite the expression as e^(ln(((2x-1)/(2x+5))^(2x+3))).

Next, we simplify the exponent by distributing it to both the numerator and denominator, resulting in e^(ln((2x-1)^(2x+3)/ (2x+5)^(2x+3))).

Now, as x approaches infinity, we can observe that the terms (2x-1) and (2x+5) dominate the expression.

Since the exponent (2x+3) is multiplied to both terms, it becomes insignificant compared to the dominant terms.

Thus, we can simplify the expression further by neglecting the exponent, resulting in e^(ln((2x-1)/(2x+5))).

Now, as x approaches infinity, the ratio (2x-1)/(2x+5) approaches 1.

Taking the natural logarithm of 1 gives us ln(1) = 0.

Finally, raising e to the power of 0 gives us e^0 = 1.

Therefore, as x approaches infinity, ((2x-1)/(2x+5))^(2x+3) evaluates to 1.

By signing up, you agree to our Terms of Service and Privacy Policy

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7