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

Question

Isaiah Allen · Accepted Answer

#lim_{x->infty}( (2x-1)/(2x+5))^{2x+3} = e^{-6}#

Making #y = 2x-1# in #f(x) =( (2x-1)/(2x+5))^{2x+3}# we get

#g(y) = (y/(y+6))^{y+4} = (y/(y+6))^y (y/(y+6))^4 #

but

# (y/(y+6))^y = (1/(1+6/y))^y = ( (1/(1+6/y))^{y/6})^6#

then

#lim_{x->infty}f(x) equiv lim_{y->infty}g(y)#

and

#lim_{y->infty}g(y) = (lim_{y->infty} (1/(1+6/y))^{y/6})^6(lim_{y->infty}( 1/(1+6/y)))^4 #

here #lim_{y->infty}( 1/(1+6/y)) =1#

From any Calculus textbook we get

#lim_{z->infty} (1/(1+1/z))^{z} = e^{-1}#

so

#lim_{y->infty}g(y) = e^{-6}#

William Abdalla · Answer

undefined 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.