Evaluate the following limits after identifying the indeterminate form. Use Hospital's rule.
d) lim x_0+ (xe^(2x) + 1)^(5/x)
e) lim x_(Pi/2)+ (1 + sec3x)^(cot3x)
Thank you!
Consider (xe^(2x) + 1)^(5/x) → 1oo
Now, ln(x) is continuous for x>0, so use that to lower the complexity
ln(xe^(2x) + 1)^(5/x) = 5/x * ln(xe^(2x) + 1)
That's now oo * 0, which we can change to 0/0 by writing it as
ln(xe^(2x) + 1)/(x/5)
Now we can apply the Rule, to get
[e^2x (2x+1)/(e^2x+1)] / (1/5) = 5
Now, that means ln(limit) = 5, and the limit is e^5
Use the same trick to change
(1 + sec3x)^(cot3x) → oo0
to get the limit = 1
To solve these limits using L'Hospital's Rule, we first need to identify the indeterminate forms in each limit.
For limit d), let's simplify the expression before identifying the indeterminate form:
lim (x -> 0+) (xe^(2x) + 1)^(5/x)
Now, we can rewrite the expression in exponential form:
lim (x -> 0+) e^[(5/x) * ln(xe^(2x) + 1)]
The indeterminate form here is 0/0.
For limit e), simplify the expression first:
lim (x -> π/2+) (1 + sec(3x))^(cot(3x))
Now, let's rewrite the expression in exponential form:
lim (x -> π/2+) e^[cot(3x) * ln(1 + sec(3x))]
Here, the indeterminate form is 0 * ∞.
Now, we can apply L'Hospital's Rule to solve each limit:
d) For the first limit, let's take the derivative of both the numerator and the denominator with respect to x:
lim (x -> 0+) [d/dx(xe^(2x) + 1)]^(5/x) / [d/dx(x)]^(5/x)
Differentiating the numerator:
lim (x -> 0+) [(1) * e^(2x) + (x) * (2e^(2x))]^(5/x) / (1)^(5/x)
Simplifying:
lim (x -> 0+) [e^(2x) + 2xe^(2x)]^(5/x) / 1
Now, the indeterminate form is 1^∞. Next, we can rewrite the expression in exponential form:
lim (x -> 0+) e^[5/x * ln(e^(2x) + 2xe^(2x))]
At this point, we can evaluate the limit by substituting x = 0 into the expression.
e) For the second limit, let's also take the derivative of both the numerator and the denominator with respect to x:
lim (x -> π/2+) [d/dx(1 + sec(3x))]^(cot(3x)) / [d/dx(cot(3x))]^(cot(3x))
Differentiating the numerator:
lim (x -> π/2+) [0 - 3sec(3x)tan(3x)]^(cot(3x)) / [−3csc^2(3x)]^(cot(3x))
Simplifying:
lim (x -> π/2+) [−3sec(3x)tan(3x)]^(cot(3x)) / [−3csc^2(3x)]^(cot(3x))
Now, the indeterminate form is 0^0. Next, let's rewrite the expression in exponential form:
lim (x -> π/2+) e^[cot(3x) * ln(−3sec(3x)tan(3x))] / [cot(3x) * ln(−3csc^2(3x))]
At this point, we can evaluate the limit by substituting x = π/2 into the expression.
Note: As both limits involve trigonometric functions, it's important to double-check the domain restrictions to make sure the limits exist.