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Find the limit, as a approaches +∞, of
Input (if a is assigned, first input purge(a)):
limit(integrate(1/(x^2),x,2,a),a,+(infinity))
Output:
Since ∫2a1/x2dx = 1/2 − 1/a, the integral
∫2a1/x2dx tends to 1/2 as a goes to infinity.
- Find the limit, as a approaches +∞, of
| ∫ | | | ⎛
⎜
⎜
⎝ | | +ln | ⎛
⎜
⎜
⎝ | | ⎞
⎟
⎟
⎠ | ⎞
⎟
⎟
⎠ | dx
|
Input (if a is assigned, first input purge(a):
limit(integrate(x/(x^2-1)+log((x+1)/(x-1)),x,2,a),a,+infinity)
Output:
Since ∫2a x/(x2−1) dx = (1/2)(ln(a2−1) − ln(3)) and
∫2aln((x+1)/(x−1))dx = ln(a+1) + ln(a−1)
+aln((a+1)/(a−1))−3ln(3), the integral
∫2ax/(x2−1) + ln((x+1)/(x−1))dx goes to infinity as a
goes to infinity.
- For an example when the integral can’t be simply evaluated, find
the limit, as a approaches 0, of
Input:
limit(int(cos(x)/x,x,a,3a),a,0)
Output:
To find this limit yourself, you can note that 1−x2/2 ≤ cos(x)
≤ 1, and so 1/x − x/2 ≤ cos(x)/x ≤ 1/x, and so
∫a3a1/x−x/2 dx ≤ ∫a3acos(x)/x dx ≤
∫a3a1/x dx, which gives you
ln(3) − 2a2 ≤ ∫a3acos(x)/x dx ≤ ln(3), and so as
a approaches 0, ∫a3acos(x)/x dx will approach ln(3).