This calculus notes will be helpful for upcoming test series for both jee and neet aspirants also this will boost your maths skills. All the best jee/ neet crackers
Limits
Definitions
Precise Definition : We say lim f (x) = L if for Limit at Infinity : We say lim f (x) = L if we can
x→a x→∞
every ε > 0 there is a δ > 0 such that whenever make f (x) as close to L as we want by taking x
0 < |x − a| < δ then |f (x) − L| < ε. large enough and positive.
“Working” Definition : We say lim f (x) = L if There is a similar definition for lim f (x) = L
x→a x→− ∞
we can make f (x) as close to L as we want by except we require x large and negative.
taking x sufficiently close to a (on either side of a)
without letting x = a. Infinite Limit : We say lim f (x) = ∞ if we can
x→a
make f (x) arbitrarily large (and positive) by taking x
Right hand limit : lim f (x) = L. This has the sufficiently close to a (on either side of a) without
x→a+
same definition as the limit except it requires x > a. letting x = a.
Left hand limit : lim f (x) = L. This has the same There is a similar definition for lim f (x) = −∞
−
x→a x→a
definition as the limit except it requires x < a. except we make f (x) arbitrarily large and negative.
Relationship between the limit and one-sided limits
lim f (x) = L ⇒ lim f (x) = lim− f (x) = L lim f (x) = lim− f (x) = L ⇒ lim f (x) = L
x→a x→a+ x→a x→a+ x→a x→a
lim f (x) 6= lim− f (x) ⇒ lim f (x)Does Not Exist
x→a+ x→a x→a
Properties
Assume lim f (x) and lim g(x) both exist and c is any number then,
x→a x→a
f (x)
lim f (x)
1. lim [cf (x)] = c lim f (x) 4. lim = x→a provided lim g(x) 6= 0
x→a x→a x→a g(x) lim g(x) x→a
x→a
h in
n
2. lim [f (x) ± g(x)] = lim f (x) ± lim g(x) 5. lim [f (x)] = lim f (x)
x→a x→a x→a x→a x→a
hp i q
3. lim [f (x)g(x)] = lim f (x) lim g(x) 6. lim n f (x) = n lim f (x)
x→a x→a x→a x→a x→a
Basic Limit Evaluations at ±∞
1. lim ex = ∞ & lim ex = 0 5. n even : lim xn = ∞
x→∞ x→− ∞ x→± ∞
b 7. n even : lim a xn + · · · + b x + c = sgn(a)∞
x→± ∞
3. If r > 0 then lim =0
x→∞ xr
8. n odd : lim a xn + · · · + b x + c = sgn(a)∞
r x→∞
4. If r > 0 and x is real for negative x
b 9. n odd : lim a xn + · · · + c x + d = − sgn(a)∞
then lim =0 x→−∞
x→− ∞ xr
Note : sgn(a) = 1 if a > 0 and sgn(a) = −1 if a < 0.
Evaluation Techniques
Continuous Functions L’Hospital’s/L’Hôpital’s Rule
If f (x)is continuous at a then lim f (x) = f (a) f (x) 0 f (x) ±∞
x→a If lim = or lim = then,
x→a g(x) 0 x→a g(x) ±∞
Continuous Functions and Composition f (x) f 0 (x)
lim = lim 0 , a is a number, ∞ or −∞
x→a g(x) x→a g (x)
f (x) is continuous at b and lim g(x) = b then
x→a
lim f (g(x)) = f lim g(x) = f (b) Polynomials at Infinity
x→a x→a
p(x) and q(x) are polynomials. To compute
Factor and Cancel p(x)
x2 + 4x − 12 (x − 2)(x + 6) lim factor largest power of x in q(x) out of
x→± ∞ q(x)
lim = lim
x→2 x2 − 2x x→2 x(x − 2) both p(x) and q(x) then compute limit.
x+6 8 3x2 − 4 x2 3 − x42
Rationalize Numerator/Denominator 3 − x42 3
√ √ √ = lim =−
3− x 3− x 3+ x x→− ∞ 5 − 2 2
lim 2 = lim 2 √ x
x→9 x − 81 x→9 x − 81 3 + x
Piecewise Function
9−x −1
= lim √ = lim √ x2 + 5
x→9 (x2 − 81)(3 + x) x→9 (x + 9)(3 + x) if x < −2
lim g(x) where g(x) =
x→−2 1 − 3x if x ≥ −2
−1 1
= =−
(18)(6) 108 Compute two one sided limits,
lim g(x) = lim x2 + 5 = 9
Combine Rational Expressions x→−2− x→−2−
lim g(x) = lim 1 − 3x = 7
1 1 1 1 x − (x + h) x→−2+ x→−2+
lim − = lim
h→0 h x+h x h→0 h x(x + h)
One sided limits are different so lim g(x) doesn’t
x→−2
1 −h −1 1 exist. If the two one sided limits had been equal
= lim = lim =− 2
h→0 h x(x + h) h→0 x(x + h) x then lim g(x) would have existed and had the
x→−2
same value.
Some Continuous Functions
Partial list of continuous functions and the values of x for which they are continuous.
1. Polynomials for all x. 6. ln(x) for x > 0.
2. Rational function, except for x’s that give 7. cos(x) and sin(x) for all x.
division by zero.
√ 8. tan(x) and sec(x) provided
3. n x (n odd) for all x. 3π π π 3π
√ x 6= · · · , − , − , , ,···
4. n x (n even) for all x ≥ 0. 2 2 2 2
9. cot(x) and csc(x) provided
5. ex for all x.
x 6= · · · , −2π, −π, 0, π, 2π, · · ·
Intermediate Value Theorem
Suppose that f (x) is continuous on [a, b] and let M be any number between f (a) and f (b). Then there exists
a number c such that a < c < b and f (c) = M .
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