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SOLUTIONS TO EXTRA PROBLEMS
- Calculate the limit, or show that it does not exist.
xlim→ 1
ex−^1 − sin(x − 1) − cos(x − 1) (ln(x))^2 We have ex^ = 1 + x + x^2 /2 + higher order terms in x so ex−^1 = 1 + (x − 1) + (x − 1)^2 /2 + higher order terms in (x − 1). From now on, H.O.T. will mean higher order terms in (x − 1). As for sin and cos, sin(x − 1) = (x − 1) − (x − 1)^3 /3 + H.O.T. and cos(x − 1) = 1 − (x − 1)^2 /2 + H.O.T.. Thus the numerator is
ex−^1 − sin(x − 1) − cos(x − 1) = 1 + (x − 1) + (x − 1)^2 / 2 − (x − 1) + (x − 1)^3 / 3 − 1 + (x − 1)^2 /2 + H.O.T. = (x − 1)^2 + (x − 1)^3 /3 + H.O.T. As for the denominator, we also have ln(x) = (x − 1) + H.O.T.
so ln(x)^2 = (x − 1)^2 + H.O.T.. Cancelling the (x − 1)^2 in the numerator and denominator, and lettin x → 1, we find that the limit is 1.
- If f (x) = ln(cos(x^2 )), what is f (4)(0)? [Hint: What’s the Taylor series for f at 0?]
Recall if T 4 is the degree four Taylor polynomial at zero then
(1) T 4 (x) = f (0) + f (1)(0)x + · · · + f (4)(0) 4! x^4. In our case, cos(x^2 ) = 1 − x^4 /2 + H.O.T. where H.O.T. means higher order terms in x. Thus ln(cos(x^2 )) = (cos(x^2 ) − 1) − (cos(x^2 ) − 1)^2 +... = (−x^4 /2 + H.O.T.) − (−x^4 /2 + H.O.T.)^2 + H.O.T. = −x^4 /2 + H.O.T.. That is, for f (x) = ln(cos(x^2 )),
(2) T 4 (x) = −x^4 / 2.
Comparing degree-four terms in (1) and (2), we find that f (4)(0)/4! = − 1 / 2
or f (4)(0) = − 12.
- Find the fourth-degree Taylor polynomial for f (x) = (^1) −^1 x 2 at a = 0.
You can do this by direct calculation. We have f ′(x) = 2 x (1 − x^2 )^2 f (2)(x) = 8 x
2 (1 − x^2 )^3
(1 − x^2 )^2
f (3)(x) = 48 x^3 (1 − x^2 )^4 +^
24 x (1 − x^2 )^3
f (4)(x) = 384 x^4 (1 − x^2 )^5
288 x^2 (1 − x^2 )^4
(1 − x^2 )^3
Then
f (0) = 1 f ′(0) = 0 f ′′(0) = 2 f ′′′(0) = 0 f ′′′′(0) = 24 = 4! It follows that T 4 (x) = 1 + 2x^2 /2 + 4!x^4 /4! = 1 + x^2 + x^4. Here’s a much easier way. The geometric series is the Taylor series for 1/(1 − x) at x = 0. It’s 1 1 − x = 1 +^ x^ +^ x
(^2) + x (^3) + x (^4) +....
Changing x to x^2 , we have 1 1 − x^2 = 1 +^ x
(^2) + x (^4) + x (^6) + x (^8) +...
so the degree-four Taylor polynomial is 1 + x^2 + x^4
- Use the -δ definition of limit to prove that
hlim→ 0
(1 + h)^3 − 1 h
We have to solve the inequality ∣∣ (1 + h)^3 − 1 h
and our solution has to be of the form |h| < δ, where δ depends only on , not h. Expanding and simplifying, we have ∣∣ 1 + 3h + 3h^2 + h^3 − 1 h
∣∣ 3 h + 3h^2 + h^3 h
3 + 3h + h^2 − 3
3 h + h^2
Now you have a quadratic in h and you’ve done a few of these before. Here’s the way we’ve done them before. Factor to get |h||3 + h| < .
We have V = b^2 h 3
By logarithmic differentiation we have dV V = 2 db b
Thus a 0.5% increase in the base would lead to approximately a 1% increase in volume. A 1% increase in height would also lead to approximately a 1% increase in volume.
- Use differentials to estimate 3
Let y = 3
x. We have dy dx =
3 x
Thus
y| 27 = 3 dy dx
272 /^3
The linear approximation to y at x = 27 is
y ∼ y| 27 + dy = y| 27 + dy dx | 27 dx
= 3 +
dx.
Setting dx = − 0 .1, we have √ (^326) .9 = 3 + 1 27
You don’t need to go any farther, but that’s about 2. 996 , and that’s correct to three decimal places. Let’s do this problem again the same way, but using slightly different notation. This is just to help you match differential notation to other notations for linear approximation. Let f (x) = x^1 /^3. The linear approximation to f at 27 is T 1 (x) = f (27) + f ′(27)(x − 27).
We have f (27) = 3 and f ′(27) =^1 3
272 /^3
So T 1 (26.9) = 3 +^1 27
