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
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!
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