Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
lim x→∞ {(5x − 4)/(5x + 3)}^5x + 1
lim x→∞ {(5x − 4)/(5x + 3)}^5x
lim x→∞ {(1 - 7/(5x+3))}^5x
Let u=5x
lim u→∞ {(1 - 7/u)}^u
I assume you can now see that this is e^-7
it is of the form inf/inf.
taking the derivatives.
it is not clear to me is the +1 in the exponent or not, I assume not.
lim= (5)/(5))*(4x)(5)
lim= 1^inf * 5=5
Why did the math book go to therapy?
Because it had too many limit issues!
To find the limit of the expression, we can first rewrite it as:
lim x→∞ [(5x − 4)/(5x + 3)]^(5x + 1)
Now let's simplify the expression by applying l'Hôpital's Rule:
lim x→∞ [(5x − 4)/(5x + 3)]^(5x + 1)
= lim x→∞ [(5x − 4)/(5x + 3)]^(5x + 1) * ln [(5x − 4)/(5x + 3)]
Now, let's evaluate the derivative of the natural logarithm:
lim x→∞ [(5x − 4)/(5x + 3)]^(5x + 1) * ln [(5x − 4)/(5x + 3)]
= lim x→∞ [(5x − 4)/(5x + 3)]^(5x + 1) * [d/dx ln ((5x − 4)/(5x + 3))]
Using the derivative of the logarithm, we get:
= lim x→∞ [(5x − 4)/(5x + 3)]^(5x + 1) * [d/dx (((5x − 4)/(5x + 3))]
= lim x→∞ [(5x − 4)/(5x + 3)]^(5x + 1) * [( ( (5(5x + 3)) - ((5x − 4)(5)) ) / (5x + 3)^2 )]
Now let's simplify this expression:
= lim x→∞ [(5x − 4)/(5x + 3)]^(5x + 1) * [( ( (25x + 15) - (25x − 20) ) / (5x + 3)^2 )]
= lim x→∞ [(5x − 4)/(5x + 3)]^(5x + 1) * [( 35 / (5x + 3)^2 )]
Since x approaches infinity, we can let the leading term dominate, meaning 35 / (5x + 3)^2 approaches 0, and thus, the limit becomes:
= lim x→∞ [(5x − 4)/(5x + 3)]^(5x + 1) * 0
= 0
Therefore, the limit of the expression is 0.
To find the limit, we can rewrite the expression and use l'Hôpital's Rule if necessary.
First, let's rewrite the expression as e raised to the power of the natural logarithm of the expression:
lim x→∞ e^ln({(5x − 4)/(5x + 3)}^5x + 1)
Now, let's apply the properties of logarithms:
lim x→∞ e^[(5x + 1) * ln((5x − 4)/(5x + 3))]
Now we have an indeterminate form of the type ∞ * 0. To apply l'Hôpital's Rule, we need to take the natural logarithm of both the numerator and denominator:
lim x→∞ [(5x + 1) * ln((5x − 4)/(5x + 3))] = ∞ * ln((5x − 4)/(5x + 3))
Now, let's simplify this expression by dividing both the numerator and denominator by x:
lim x→∞ [ln((5 - 4/x)/(5 + 3/x))] = ∞ * ln((5 - 4/x)/(5 + 3/x))
Now we have an indeterminate form of the type ∞ * ln(1/1). Since ln(1) is equal to 0, the expression simplifies to:
lim x→∞ [ln(1)] = ∞ * 0 = 0
Therefore, the limit of the given expression as x approaches infinity is 0.
To evaluate the given limit, we can first simplify the expression inside the limit. Then, we can try to apply l'Hôpital's Rule if necessary.
First, let's simplify the expression:
lim x→∞ {(5x − 4)/(5x + 3)}^5x + 1
Since we're dealing with an exponential function, let's try taking the natural logarithm of the expression to simplify it:
ln[lim x→∞ {(5x − 4)/(5x + 3)}^5x + 1]
Using the properties of logarithms, we can bring down the exponent to the front:
lim x→∞ (5x + 1) ln[(5x − 4)/(5x + 3)]
Next, we can use algebraic manipulation to simplify the expression inside the logarithm:
lim x→∞ (5x + 1) ln[(5 − 4/x)/(5 + 3/x)]
As x approaches infinity, both 4/x and 3/x approach 0. Therefore, we can simplify further:
lim x→∞ (5x + 1) ln[(5/5)]
Since ln(1) = 0, the expression further simplifies:
lim x→∞ (5x + 1) (0)
Now, we can evaluate the limit:
lim x→∞ 0
The limit of a constant function is equal to the constant itself. Therefore, the limit is:
0