Find the limit as x->infinity of ((3x+2)/(3x+4))^(3x+1)
multiply numerator and denominator by
(1/x)^(3x+1)
Lim ((3+2/x)/(3+4/x))^(3x+1)
so as x>>inf this reduceds to
((3/3)^3x+1) and as x>>inf
= 1
Thank You. This works,and easier I also did it by finding the ln of the expression
(3x+1)[ln(3x-2)-ln(3x+4)]
changed to [ln(3x-2)-ln(3x+4)]/(1/(3x+1)
then used L'Hospital Rule to get lim=0
so since ln y = 0, the e^0=1
To find the limit as x approaches infinity of an expression, we can look at the highest power of x in the numerator and denominator. In this case, the highest power of x is (3x+1).
To evaluate the limit, divide every term in the expression by x^(3x+1). This will help us simplify the expression and make it easier to find the limit.
((3x+2)/(3x+4))^(3x+1)
= ((3x+2)/(3x+4)) * (1/x)^(3x+1)
= [(3x+2)/(3x+4)] * [1/(x^(3x)) * 1/x]
Now let's evaluate each part separately:
- Limit of (3x+2)/(3x+4) as x approaches infinity:
Since the highest power of x is x, we can ignore the constant terms (2 and 4). Divide every term by x, and we get:
(3+2/x)/(3+4/x)
As x approaches infinity, the terms with 2/x and 4/x become negligible, so we are left with:
3/3
= 1
- Limit of (1/x)^(3x) as x approaches infinity:
To evaluate this, rewrite it as e^[(3x)*ln(1/x)] since ln(1/x) is the natural logarithm of x.
= e^[(3x)*ln(1)-ln(x)]
= e^[-3x*ln(x)]
As x approaches infinity, both -3x and ln(x) approach infinity. However, their product -3x*ln(x) is indeterminate. We can apply L'Hôpital's rule to find the limit of this part.
Differentiate both the numerator and denominator with respect to x:
(-3*ln(x) - 3)/(1/x)
Simplifying further:
= (-3*ln(x) - 3)*x
= -3x*ln(x) - 3x
As x approaches infinity, -3x approaches negative infinity, whereas ln(x) approaches positive infinity. Their product -3x*ln(x) also approaches negative infinity.
Multiplying the two limits together:
1 * (-3x*ln(x)) = -3x*ln(x)
Therefore, the limit of ((3x+2)/(3x+4))^(3x+1) as x approaches infinity is -3x*ln(x).