Calculus/calc

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Expert: - 6/12/2008


Question
hello! my question is, using l'hopitals rule what is the limit of (1+3x)^cscx as x approaches zero from the right? if you could tell me the steps that would be amazing thanks!

Answer
This problem needs to be rewritten before applying L'Hospital's rule.

Any function f(x) can be rewritten as e^(ln(f(x)).  
The function ln(f(x)^n) can be rewritten as n*ln(f(x).
To apply L'Hospital's rule, we need a quotient, so n*ln(f(x)) can be rewritten as ln(f(x)/(1/n).

What we have is the e^(ln(1+3x)/(1/csc(x))), or e^(ln(1+3x)/sin(x)).

Apply L'Hospital's rule to the exponet and you get
e^(ln'(1+3x)/sin'(x)) = e^(3/(1+3x))/(cos(x)).

As x->0, 3/(1+3x)->3 and cos(x)->1, so the fraction goes to 3, and the answer is e^3.  

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