# How do you find the Limit of #1/(lnx)# as x approaches infinity?

0

By signing up, you agree to our Terms of Service and Privacy Policy

To find the limit of 1/(lnx) as x approaches infinity, we can use the concept of L'Hôpital's Rule. By applying this rule, we differentiate the numerator and denominator separately and then take the limit as x approaches infinity.

Differentiating the numerator, we get 0, as the derivative of a constant is always 0.

Differentiating the denominator, we use the chain rule. The derivative of ln(x) is 1/x.

Taking the limit as x approaches infinity, we have 0/(1/x), which simplifies to 0.

Therefore, the limit of 1/(lnx) as x approaches infinity is 0.

By signing up, you agree to our Terms of Service and Privacy Policy

- For what values of x, if any, does #f(x) = 1/((x-3)(x+4)(e^x-x)) # have vertical asymptotes?
- How do you find the limit of #(sqrt(x+5)-3)/(x-4)# as #x->4#?
- How do you find the limit of #(x^2-3x+2)/(x^3-4x)# as x approaches 2 from the right, as x approaches -2 from the right, as x approaches 0 from the left, and as x approaches 1 from the right?
- How do you use the epsilon delta definition of limit to prove that #lim_(x->1)(x+2)= 3# ?
- What is the limit of #(sqrt(x^2+4x+1)-x) # as x goes to infinity?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7