Limit as n approaches infinity:
[(n + 3)/(n + 1)]^n
How do I even start this?
I would divide the polynomials...
lim (1+ 2/n + ...)^n= 1
Answer is e^2.
lim (n -> infinity) of [(n+3)(n+1)]^n
= lim (n -> infinity) of [1 + 2/(n+1)]^n
e = lim (x -> infinity) of [1 + 1/x]^x
e^2 = lim (x -> infinity) of [1 + 2/x + 1/x^2]^x
To find the limit as n approaches infinity for the expression [(n + 3)/(n + 1)]^n, we can utilize the concept of limits and the properties of exponential functions.
Step 1: Rewrite the expression using exponential notation.
[(n + 3)/(n + 1)]^n can be rewritten as [(n + 3)/(n + 1)]^n = [(1 + 3/n)/(1 + 1/n)]^n.
Step 2: Simplify the expression inside the brackets.
We can simplify the brackets by dividing both the numerator and the denominator by n.
[(1 + 3/n)/(1 + 1/n)] = [(1/n)(1 + 3/n)] / [(1/n)(1 + 1/n)].
Step 3: Evaluate the limits separately.
Now let's evaluate the limits of both the numerator and the denominator separately as n approaches infinity.
- For the numerator: lim(n → ∞)[(1/n)(1 + 3/n)] = (0)(1) = 0.
- For the denominator: lim(n → ∞)[(1/n)(1 + 1/n)] = (0)(1) = 0.
Step 4: Rewrite the expression using the limits.
[(1/n)(1 + 3/n)] / [(1/n)(1 + 1/n)] = 0/0.
Step 5: Apply L'Hôpital's Rule.
L'Hôpital's Rule allows us to compute the limit of an indeterminate form (such as 0/0) by taking the derivative of the numerator and the denominator repeatedly until we get a non-indeterminate form.
Take the derivative of the numerator and the denominator with respect to n:
- Derivative of the numerator: d/dn[(1/n)(1 + 3/n)] = [(0)(1 + 3/n) - (1/n^2)(1 + 3/n)] = -3/n^2.
- Derivative of the denominator: d/dn[(1/n)(1 + 1/n)] = [(0)(1 + 1/n) - (1/n^2)(1 + 1/n)] = -1/n^2.
Now compute the limit again with the derivatives:
lim(n → ∞)(-3/n^2)/(-1/n^2) = lim(n → ∞)(3/-1) = -3.
Therefore, the limit as n approaches infinity for the expression [(n + 3)/(n + 1)]^n is -3.