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MATH 100 Winter 2015 midterm exam solutions
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MATH 100 Winter 2015 midterm exam solutions
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Solutions for Math 100 2015 Midterm1. Each limit exists and is given by
(a) p p p
x 1 ( x 1) ( x + 1)
lim p = lim p p
x!1 1 x2 x!1 (1 x) (1 + x) ( x + 1)
1 x 1 1 p
= p lim p = p lim 1 x = 0:
2 2 x!1 1 x 2 2 x!1
(b)
p p sin (1=x)
lim 1 + x2 tan (1=x) = lim jxj 1 + 1=x2
x! 1 x! 1 cos (1=x)
p sin (1=x) sin (1=x)
= lim x 1 + 1=x2 = lim = 1:
x! 1 cos (1=x) x! 1 1=x
2. For f (x) to be di¤erentiable at x = =2, we require that f (x) be contin-
uous at x = =2 and that
+
f0 = f0 :
2 2
From continuity we have
a + 4 cos ( =2) = 1 + b sin ( ) =) a = 1:
And from the condition on the derivatives we have
4 sin =2 = 2b cos =) b = 2:
3. The derivatives are given by
(a) y 0 = 2xesin x + x2 cos (x) esin x ;
sech2 x
(b) y 0 = ;
2 (1 + tanh x)
y + xy 0 1 + y0
(c) q = q ;
2 2
(xy) 1 1 (x + y)
(d) y 0 = 10x9 + ln (10) 10x :
4. The distance to the origin, denoted by s, is given by
ds dx dy
s2 = x2 + y 2 =) s =x +y :
dt dt dt
When t = =2
x=1 cos =2 = 1,
1
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