a.) lim x->∞ ln[(2e^x) + (e^-x)] - ln[(e^x) - (e^-x)] b.) lim x->0^+ [tan^-1(lnx)] c.) lim x-> -∞ [cos^-1(2-x^2 / x^2+3x)] d.) lim x->0^+ [lnx / tanh^-1(x-1)] e.) lim x->0^- [(cos(3x))^2/x]

Question

Find the indicated limits.

Answer 1

Part a: Divide the numerator & denominator by e^x and the answer will be apparent. Remember:

ln a - ln b = ln(a/b)...threre is your fraction.

Part b: What is the tan (-pi/2)? As x->0, ln x -> -∞ so that you are looking for an angle x for which tan(x)=-∞

Part c: As x->-∞ both the numerator and denominator approach 1, but the quotient is always greater than -1.

What does that tell you about the limit?

Part d: The numerator is unbounded while the denominator is bounded.

Part e: The denominator is bounded and the denominator goes to 0.

I will NOT work these problems for you completely; I have given you sufficient clues to finish the problem by yourself if you will think about what you need to do.