Posted by Tommy on Thursday, April 26, 2012 at 10:33pm.
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
Related Questions
calculus - Evaluate the following limits after identifying the indeterminate ...
Calculus - L'Hospital's Rule Lim(lnx)^(x-1) x->1+ This is what I ...
math - i need some serious help with limits in pre-calc. here are a few ...
calculus - Using L'Hôpital's rule, evaluate lim of xe^(-x) as x ...
calculus - Evaluate the integral of xe^2x(dx) A. (1/6)(x^2)(e^3x)+C B. (1/2)(xe^...
Maths Calculus Derivatives & Limits - Using the definition of the ...
Calculus - Evaluate the limit using L'Hospital's rule if necessary. lim ...
Calculus - Find the following limits algebraically or explain why they don...
Calculus - F(x)=|x-5| evaluate the following limits lim x->5- ( f(x)-f(5...
Calculus..more help! - I have a question relating to limits that I solved lim(x...
For Further Reading