| |
You are here: Experts > Teens > Homework/Study Tips > Calculus > calculus
Expert: Ahmed Salami - 10/28/2009
Question
what happens if you try to use l'hopital's rule to evaluate limit as x goes to infinity of x divided by the square root of x^2+1 ???
please help me...thanks!!!
Answer
Hi Angel,
L'Hopital's rule is used to evaluate a limit when direct substitution results in the indeterminate forms 0/0 or ∞/∞.
For x/√(x² + 1), direct substitution results in ∞/∞
Using L'Hopital's rule
lim x/√(x² + 1) = lim 1/[1/2√(x² + 1)]
= lim 2√(x² + 1)]
and the method fails.
To solve this limit, divide top and bottom by x to get
x/√(x² + 1) = (x/x) / [√(x² + 1)/x]
= (x/x) / [√(x² + 1)/√x²]
= 1 / √(1 + 1/x²)
As x → ∞, 1/x² → 0 and the function approaches 1 / √(1 + 0) = 1
Now, you can see why the method would fail. By algebraic manipulation, expressions can be made to give indeterminate forms after direct substitution e.g 0/0 by multiplying the expression by x/x. To show you, as in this problem
lim 1/√(1 + 1/x²) = lim 1/√(1 + 1/x²).(x/x)
= lim x/√(x² + 1) by simply working backwards
But of course L'Hopital's rule doesnt apply to lim 1/√(1 + 1/x²) in the first place, and the rule is only effective for particular non-cyclic problems.
Regards
Add to this Answer Ask a Question
|
|
