Some Computational Problems on Limits and Continuity

Aditya Ghosh

1. Suppose that lim (f (x) + g(x)) = 2 and lim (f (x) − g(x)) = 1. Is it necessary that
x→a x→a
lim f (x)g(x) exists? If yes, can you calculate that limit?
x→a
p
1 − cos 2(x − 1) x3 − x2 log x + log x − 1
2. Find the following limits: (i) lim , (ii) lim ,
√ √ √ x→1 x−1 x→1 x2 − 1
3
x+ x+x x−3
(iii) lim
x→1 x−1

5x − 4, if 0 < x ≤ 1,
3. Define a function f (x) = Find out whether lim f (x) exists.
4x3 − 3x, if 1 < x < 2. x→1
Is f continuous at x = 1?
3x2 + ax + a + 1
4. Let f (x) = . Determine the value(s) of a for which (i) lim f (x) exist,
x2 + x − 2 x→1
(ii) lim f (x) exists. Do you have any conclusion? Also calculate those two limits.
x→−2

bx/2c
5. Determine lim , where b·c is the box/floor function.
x→π/2 log(sin x)

xp+1 − (p + 1)x + p
6. Determine lim where p is a positive integer.
x→1 (x − 1)2
p √
sin2 x − sin2 y sin(π cos2 x) 1 − sin 2x
7. Evaluate: (i) lim , (ii) lim , (iii) lim .
x→y x2 − y 2 x→0 x2 x→π/4 π − 4x
√  √ √ 
8. Calculate the following limits: lim x2 + x − x and lim x2 + x + 1 − x2 + 1 .
√ x→∞ x→∞
Also find lim n sin(2π 1 + n2 ).
n→∞

9. Calculate the following limits:

10. Find lim− f (x) and lim+ f (x) for the following function:
x→0 x→0
 2
tan ({x})/(x2 − [x]2 ), if x > 0,
f (x) = 1, p for x = 0,
1/ {x} cot{x}, for x < 0.


Here [x] is the floor function and {x} = x − [x].

1
11. If α, β be the roots of the quadratic equation ax2 + bx + c = 0, prove that

1 − cos(ax2 + bx + c) 1
lim 2
= (b2 − 4ac).
x→β (x − β) 2

tan({x} − 1) sin({x})
12. Find whether the limit lim exists.
x→0 {x}({x} − 1)
log(2 + x) − x2n sin x
13. Discuss the continuity of the function f (x) = lim at x = 1.
n→∞ 1 + x2n
 1 x
The next few problems will involve the concept of e := lim 1 + . This number e is
x→∞√ x
a real number, just like any other real number, say 2, −1, 2/3, 3, or π/2. Then you might
wonder, why is e so special? Actually it is the function exp(x) := ex that makes e so special.
Here is a note that attempts to give some insights about different results related to e that we
usually use in a first course of Calculus. In particular, we assume the following limit without
proof:
ex − 1
lim = 1.
x→0 x
A proof of this is given in the above note, you may read that once you are ready. (The way I
treated it in the note requires the knowledge of Fundamental theorem of Integral Calculus).
Assuming the above limit (and the above definition of e), you can prove the following limits:

log(1 + x) ax − 1
lim = 1, lim = loge a, lim (1 + x)1/x = e.
x→0 x x→0 x x→0

 
Result. If lim f (x) = 0, then lim (1 + f (x))g(x) = exp lim f (x)g(x) . (Prove it yourself!)
x→a x→a x→a
This result is often used when we know that lim a(x) = 1, and we wish to find lim a(x)b(x) .
 x→a   x→a
b(x)
We just do this: lim a(x) = exp lim b(x) log a(x) = exp lim (a(x) − 1)b(x) .
x→a x→a x→a

Key idea: Whenever you have a(x)b(x) , take log.

px − q x  π 1/x
14. lim 18. lim tan +x
x→0 r x − sx x→0 4
1/x
log(1 + 2h) − 2 log(1 + h) ax + b x + c x

15. lim 19. lim
h→0 h2 x→0 3
 a bx  x
16. lim 1 + 1 1
x→∞ x 20. lim sin + cos
x→∞ x x
 x + 6 x+4 2
17. lim 21. lim (cos x)cot x
x→∞ x + 1 x→0