Dayaan M. answered • 09/11/26
Earned A’s in Calc 1/AB & Calc 2/BC | 5 Years of Tutoring Experience
You can solve for δ exactly here instead of guessing and checking. The answer turns out to be δ = arctan(1/√M).
What the definition is asking for
Definition 6 says limx→π/2 tan2x = ∞ means: for every M > 0 there is a δ > 0 such that
0 < |x − π/2| < δ ⇒ tan2x > M
So your job is to find how close x has to be to π/2 to force tan2x above the given height M.
Solving for δ
Substitute x = π/2 + t, so that |t| = |x − π/2| is exactly the distance you're controlling. Using the identity tan(π/2 + t) = −cot t,
tan2x = cot2t = 1 / tan2t
Now just solve the inequality:
1 / tan2t > M ⇔ tan2t < 1/M ⇔ |tan t| < 1/√M ⇔ |t| < arctan(1/√M)
That last step is valid because tan is increasing and positive on (0, π/2). So take
δ = arctan(1/√M)
(a) M = 1000
δ = arctan(1/√1000) = arctan(0.0316228) ≈ 0.0316
(b) M = 10,000
δ = arctan(1/100) = arctan(0.01) ≈ 0.0100
Check that it's really working
Take M = 1000 and step just inside the boundary, t = 0.031612. Then tan2(π/2 + t) ≈ 1000.002, which clears 1000. Step just outside, t = 0.031613, and you get ≈ 999.998 — below M. So this δ is not merely one that works, it's the largest one that works. Any smaller δ is also a correct answer to the problem; the definition only asks you to produce one.
Two things worth noticing
First, multiplying M by 10 only shrank δ by about a factor of √10 ≈ 3.16, not by 10. That's because tan2 blows up like 1/t2 near π/2, so you need |t| < 1/√M — the square root is doing the work.
Second, for small angles arctanθ ≈ θ, which is why δ ≈ 1/√M is such a good approximation: 0.0316122 vs 0.0316228, and 0.0099997 vs 0.0100000. If your textbook's answer key says 0.0316 and 0.01, that's the same result — they just used the approximation.
One note on the problem statement: the "= 8" in your title is a garbled ∞. The limit is infinity, which is why this uses the M-version of the definition rather than the usual ε-δ one.
