# Calculus

Find the limit: lim x-> 2 ln(x/2)/ (x^2−4)

Can someone help me with this?

1. 👍
2. 👎
3. 👁
1. I thought that it was 1/4 but that's now the answer....

1. 👍
2. 👎
2. What I thought that I had to do was to take the derivative, (2/x)/2x and then plug in 2... and that's how I got 1/4, but that's not the answer, what do I have to do?

1. 👍
2. 👎
3. D'hôpital's rule works here.
The expression is
ln(x/2)/ (x^2−4)

not
2ln(x/2)/ (x^2−4)
as it appears in the post.

The leading 2 belongs to the limit of x.

Differentiate both top and bottom with respect to x:
Lim (1/x) / (2x)
= lim 1/(2x²)
=1/(2(2)²)
=1/8

1. 👍
2. 👎

## Similar Questions

1. ### Calculus

The area A of the region S that lies under the graph of the continuous function is the limit of the sum of the areas of approximating rectangles. A = lim n → ∞ [f(x1)Δx + f(x2)Δx + . . . + f(xn)Δx] Use this definition to

2. ### Calculus

Let f be a function defined for all real numbers. Which of the following statements must be true about f? Which might be true? Which must be false? Justify your answers. (a) lim of f(x) as x approaches a = f(a) (b) If the lim of

1. Evaluate: lim x->infinity(x^4-7x+9)/(4+5x+x^3) 0 1/4 1 4 ***The limit does not exist. 2. Evaluate: lim x->infinity (2^x+x^3)/(x^2+3^x) 0 1 3/2 ***2/3 The limit does not exist. 3. lim x->0 (x^3-7x+9)/(4^x+x^3) 0 1/4 1 ***9 The

4. ### calc urgent

Note that f is continuous on (−∞, 6) and (6, ∞). For the function to be continuous on (−∞, ∞), we need to ensure that as x approaches 6, the left and right limits match. First we find the left limit. lim x→6− f(x)

1. ### Calculus Limits

Question: If lim(f(x)/x)=-5 as x approaches 0, then lim(x^2(f(-1/x^2))) as x approaches infinity is equal to (a) 5 (b) -5 (c) -infinity (d) 1/5 (e) none of these The answer key says (a) 5. So this is what I know: Since

2. ### calculus

1) find the indicated limit, if it exist? a) lim x->-2 (x^2 -9)/(x^2+x-2) b) lim x -> -∞ √(ax^2+bx+c)/dx + e, where a > 0, b,c,d, and e are constant.

3. ### Check my CALCULUS work, please! :)

Question 1. lim h->0(sqrt 49+h-7)/h = 14 1/14*** 0 7 -1/7 Question 2. lim x->infinity(12+x-3x^2)/(x^2-4)= -3*** -2 0 2 3 Question 3. lim x->infinity (5x^3+x^7)/(e^x)= infinity*** 0 -1 3 Question 4. Given that: x 6.8 6.9 6.99 7.01

4. ### Math

Use graphing utility to graph function and estimate the limit. Use a table to reinforce your conclusion. Then find the limit by analytic methods. lim (sq root of (x+2) - sq root of 2) / x x->0 Thanks!

1. Evaluate: lim x->infinity(x^4-7x+9)/(4+5x+x^3) 0 1/4 1 4 The limit does not exist. 2. Evaluate: lim x->infinity (2^x+x^3)/(x^2+3^x) 0 1 3/2 2/3 The limit does not exist. 3. lim x->0 (x^3-7x+9)/(4^x+x^3) 0 1/4 1 9 The limit does

2. ### limits

find the limit of lim x--->0^+ x/ln(x) lim x----> inf (1+4/x)^x

3. ### Math

Evaluate the limit using the appropriate Limit Law(s). (If an answer does not exist, enter DNE.) lim t → −1 (t2 + 1)^4(t + 3)^5

4. ### Calculus

Find the limit. lim 5-x/(x^2-25) x-->5 Here is the work I have so far: lim 5-x/(x^2-25) = lim 5-x/(x-5)(x+5) x-->5 x-->5 lim (1/x+5) = lim 1/10 x-->5 x-->5 I just wanted to double check with someone and see if the answer is