Solving L'Hospital's Rule: Cosx/x^2

  • Thread starter KLscilevothma
  • Start date
In summary, the limit of (cosx)(1/x^2) as x approaches 0 is 1/sqrt(e) or e^(-1/2). This can be found by using the L'Hospital's rule and taking the natural log of the function. The final answer can also be simplified to e^(-1/2) using basic algebraic rules.
  • #1
KLscilevothma
322
0
Q1) lim (cosx)(1/x^2)
x-->0

I tried to find

lim (1/x^2) ln cos x
x-->0

but got stuck after differentiate it several times.

Thanks in advance
 
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  • #2
L'Hospital is quite complicated to be applied here...

(cosx)^(1/x^2)=[1-(sinx)^2]^(1/(2*(x^2)))=

={[1-(sinx)^2]^[-1/((sinx)^2)]}^(-((sinx)^2)/(2*(x^2)))

lim {[1-(sinx)^2]^[-1/((sinx)^2)]}^(-((sinx)^2)/(2*(x^2)))=
x->0

=lim {[1-(sinx)^2]^[-1/((sinx)^2)]}^lim(-((sinx)^2)/(2*(x^2)))=
x->0____________________________x->0
=e^(-1/2)=1/sqrt(e);

I used sinx/x -> 1 as x->0 and

lim (1+Xn)^(1/Xn)=e, where
x->0

lim Xn = 0, Xn >0 or Xn <0...
n->infinity

I must remember you L'Hospital's rule :

lim f(x)/g(x)=lim f'(x)/g'(x),
x->a_______x->a

and
lim f(x)=0 or infinity
x->a
lim g(x)=0 or infinity
x->a
...
 
Last edited:
  • #3
Thank you bogdan!

I just got the answer before viewing your thread. I made a very silly mistake when doing this question at the very beginning.

Here is my approach.
lim (cosx)(1/x2)
x-->0

Let y=(cosx)(1/x2)
take natural log on both sides
ln y = (1/x2) ln cos x

Lim (1/x2) ln cos x
x-->0

=Lim ln(cosx)/x2 (which is an indeterminate form of 0/0)
x-->0

=Lim -sinx/(2xcosx)
x-->0

=Lim -(sinx)/x * 1/(2cosx)
x-->0

=-1/2

Therefore
lim (cosx)(1/x^2) = e-1/2
x-->0
 
  • #4
Very nice...and simple...
 

1. What is L'Hospital's Rule?

L'Hospital's Rule is a mathematical principle used to evaluate limits involving indeterminate forms, particularly in calculus. It states that if the limit of a function f(x) as x approaches a certain value is equal to 0/0 or ∞/∞, then the limit is equal to the limit of the derivative of f(x) divided by the derivative of g(x), where g(x) is the denominator of the original function.

2. How is L'Hospital's Rule applied to Cosx/x^2?

In this case, the original function is cosx/x^2, and both the numerator and denominator approach 0 as x approaches 0. To apply L'Hospital's Rule, we take the derivative of both the numerator and denominator, which gives us -sinx and 2x, respectively. Then, we evaluate the limit of -sinx/2x as x approaches 0, which is equal to -1/2.

3. Are there any limitations to using L'Hospital's Rule?

Yes, there are several limitations to using L'Hospital's Rule. First, the rule can only be applied to limits with indeterminate forms, as mentioned earlier. Additionally, the function must be differentiable in the given interval. Also, the limit must exist in order for the rule to be applied. Finally, L'Hospital's Rule may not always provide the correct answer, so it should be used with caution and in conjunction with other methods.

4. Can L'Hospital's Rule be applied multiple times?

Yes, L'Hospital's Rule can be applied multiple times as long as the resulting limit is still an indeterminate form. However, it is important to note that applying the rule multiple times may not always provide the correct answer and should be used with caution.

5. Are there any other methods for solving limits besides L'Hospital's Rule?

Yes, there are several other methods for solving limits, such as substitution, factoring, and using algebraic manipulation. These methods should be used when L'Hospital's Rule cannot be applied or when the function is not differentiable in the given interval. It is important to be familiar with all methods and choose the most appropriate one for each specific limit problem.

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