1
We see that both the numerator and the denominator has the limit 0 so we try
L’Hospital’s rule. Set f (x) = sin(x2 ) and g(x) = 1 − cos(x). Then f 0 (x) =
2x cos(x2 ) and g 0 (x) = sin(x). We see that we still have an ideterminate form
of the type 00 so we try L’Hospital again. We find that f 00 (x) = 2 cos(x2 ) −
4x2 sin(x2 ) and g 00 (x) = cos(x). Now we can compute the limit. Taking it all
together we get:
sin(x2 )
2x cos(x2 )
= lim
=
x→0 1 − cos(x)
x→0
sin(x)
2
2 cos(x2 ) − 4x2 sin(x2 )
= = 2.
lim
x→0
cos(x)
1
lim
Answer: The limit is 2.
2
We take the derivative of both sides with respect to x. We get
y 0 + 2yy 0 = ex ⇔ (1 + 2y)y 0 = ex ⇔ y 0 =
ex
.
1 + 2y
We use the assumption that ex = y + y 2 to rewrite the expression for y 0 :
y0 =
Answer:
dy
dx
=
y + y2
.
1 + 2y
y+y 2
1+2y .
3
√
We use the change of variables u = x + 1. We see that x = u2 − 1 and that
dx = 2udu, using the standard way of writing change of variables.
Z
2
x
√
dx =
x
+1
0
3
√3
u
2
−u
=2
3
1
Z √3
u2 − 1
2udu = 2
(u2 − 1) du =
u
1
1
!
√
√
3 3
1
4
− 3 −2
−1 = .
3
3
3
√
Z
3
Answer: The integral is equal to 43 .
4
We rewrite the equation:
y 0 = y 2 · sin(x) ⇔
y=
Since y(0) =
1
y0
= sin(x) ⇔ − = − cos(x) + C ⇔
2
y
y
1
.
cos(x) − C
1
2
we must have that
1
1
1
=
=
⇔ C = −1.
2
cos(0) − C
1−C
Answer: The solution is y(x) =
1
cos(x)+1 .
5
√
We use the root test. Set an = (−1)n en−
n n
x . Then
√
1
lim |an | n = lim e1−
n→∞
lim e
n→∞
n→∞
1− √1n
n
n
|x| =
|x| = e1−0 |x| = e|x|.
By the ratio test the series converges if e|x| < 1 and diverges if e|x| > 1.
The condition e|x| < 1 is equivalent to |x| < 1e .
Answer: The radius of convergence is 1e .