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CONCEPT OF LIMIT

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S0NAWANE ACADEMY
MATHS COACHING
CHAPTER- limits
JOB SHEET NO- -01
DEFINITION OF LIMIT:
lim 𝑓(π‘₯) exists iff
π‘₯→π‘Ž
lim
lim
𝑓(π‘₯) =
𝑓(π‘Ž − β„Ž) exists,
π‘₯ → π‘Ž−
β„Ž→0
lim
lim
(2) R.H.L.=Right hand limit= 𝑓(π‘Ž+ ) =
𝑓(π‘₯) =
𝑓(π‘Ž + β„Ž) exists,
π‘₯ → π‘Ž+
β„Ž→0
lim
lim
(3)
𝑓(π‘₯).
− 𝑓(π‘₯) =
π‘₯→π‘Ž
π‘₯ → π‘Ž+
lim
lim
If lim 𝑓(π‘₯) exists then lim 𝑓(π‘₯) =
𝑓(π‘₯) =
𝑓(π‘₯).
π‘₯→π‘Ž
π‘₯→π‘Ž
π‘₯ → π‘Ž−
π‘₯ → π‘Ž+
(1) L.H.L.=Left hand limit = 𝑓(π‘Ž− ) =
Q.1
1] Evaluate: lim 𝑓(π‘₯) where 𝑓(π‘₯) =
π‘₯→π‘Ž
π‘₯ 2 −4
at π‘₯ = 2
π‘₯−2
|π‘₯−4|
, π‘₯≠4
2] Evaluate: lim 𝑓(π‘₯) where 𝑓(π‘₯) = { π‘₯−4
π‘Žπ‘‘ π‘₯ = 4
π‘₯→π‘Ž
0,
π‘₯=4
1 + π‘₯2, 0 ≤ π‘₯ ≤ 1
3]Evaluate: lim 𝑓(π‘₯) where 𝑓(π‘₯) = {
π‘Žπ‘‘ π‘₯ = 1
π‘₯→π‘Ž
2 − π‘₯,
π‘₯>1
π‘₯−|π‘₯|
, π‘₯≠0
4]If 𝑓(π‘₯) = { π‘₯
show that lim 𝑓(π‘₯) does not exist.
π‘₯→0
2,
π‘₯=0
5π‘₯ − 4, 0 < π‘₯ ≤ 1
5]If 𝑓(π‘₯) = { 3
show that lim 𝑓(π‘₯) exists.
π‘₯→1
4π‘₯ − 3π‘₯, 1 < π‘₯ < 2
6] Discuss the existence of each of the following limits.
1
1
(i)
lim π‘₯
(ii) lim |π‘₯|
π‘₯→0
π‘₯→0
π‘π‘œπ‘ π‘₯,
π‘₯>0
7] Let 𝑓(π‘₯) = {
find the value of constant k ,
π‘₯ + π‘˜,
π‘₯<0
given that lim 𝑓(π‘₯) exists.
π‘₯→0
8]Let 𝑓(π‘₯) be function defined by
4π‘₯ − 5, π‘₯ ≤ 2
𝑓(π‘₯) = {
find 𝛼 if lim 𝑓(π‘₯) exists.
π‘₯ − 𝛼,
π‘₯>2
π‘₯→2
π‘šπ‘₯ 2 + 𝑛, π‘₯ < 0
9] If 𝑓(π‘₯) = {𝑛π‘₯ + π‘š,
0≤π‘₯≤1
3
𝑛π‘₯ + π‘š, π‘₯ > 1
For what values of integers m and n does the
limits lim 𝑓(π‘₯) and lim 𝑓(π‘₯) exists.
π‘₯→0
π‘₯→1
[NCERT]
LIMITS
|π‘₯| + 1, π‘₯ < 0
π‘₯ = 0 for what value of a does lim 𝑓(π‘₯)exist ? [NCERT]
10] If 𝑓(π‘₯) = { 0,
π‘₯→0
|π‘₯| − 1, π‘₯ > 0
π‘Ž + 𝑏π‘₯, π‘₯ < 1
11] Suppose 𝑓(π‘₯) = { 4,
π‘₯=1
𝑏 − π‘Žπ‘₯, π‘₯ > 1
And if lim 𝑓(π‘₯) = 𝑓(1) . what are possible values of a and b?
[NCERT]
π‘₯→1
JEE MAIN LEVEL EXAMPLES
(1) Find left hand and right hand limits of the greatest integer function
𝑓(π‘₯) = [π‘₯]= greatest intger less than or equal to , at π‘₯ = π‘˜ ,
where π‘˜ is an ineteger. Also show that lim 𝑓(π‘₯) does not exists.
π‘₯→π‘˜
2] Prove that lim+[π‘₯] = [π‘Ž] for all ∈ 𝑅 , where [. ] denotes the greatest integer function.
π‘₯→π‘Ž
𝑒 1/π‘₯ −1
3] Show that lim 𝑒 1/π‘₯ +1 does not exist.
π‘₯→0
4] If 𝑓is an odd function and if lim 𝑓(π‘₯) exists. Prove that this limit must be zero.
π‘₯→0
5] If 𝑓is an even function then prove that lim− 𝑓(π‘₯) = lim+ 𝑓(π‘₯) exists.
π‘₯→0
π‘₯→0
HOME WORK
CET / BOARD LEVEL
π‘₯
1. Show that lim |π‘₯| does not exist.
[NCERT]
π‘₯→0
2π‘₯ + 3, π‘₯ ≤ 2
2. Find k so that lim 𝑓(π‘₯) may exist, where (π‘₯) = {
.
π‘₯ + π‘˜, π‘₯ > 2
π‘₯→2
1
3. Show that lim π‘₯ does not exist.
π‘₯→0
4. Let 𝑓(π‘₯) be a function defined by 𝑓(π‘₯) =
3π‘₯
{|π‘₯||2π‘₯
, π‘₯≠0
0,
π‘₯=0
>0
. Prove that lim 𝑓(π‘₯) does not exist.
<0
π‘₯→0
>0
prove that lim 𝑓(π‘₯) does not exist.
<0
π‘₯→2
4,
if π‘₯ > 1
7. Find lim 𝑓(π‘₯) where 𝑓(π‘₯) = {
π‘₯ + 1, if π‘₯ < 1
π‘₯→1
2π‘₯ + 3, π‘₯ ≤ 0
8. If 𝑓(π‘₯) = {
find lim 𝑓(π‘₯) and lim 𝑓(π‘₯)
[NCERT]
3(π‘₯ + 1), π‘₯ > 0
π‘₯→0
π‘₯→1
π‘₯ 2 − 1, π‘₯ ≤ 1
9. Find lim 𝑓(π‘₯), if (π‘₯) = { 2
.
[NCERT]
π‘₯→1
−π‘₯ − 1, π‘₯ > 1
|π‘₯|
, π‘₯≠0
10. Evaluate lim 𝑓(π‘₯), where 𝑓(π‘₯) = { π‘₯
[NCERT]
π‘₯→0
0, π‘₯ = 0
11. Let π‘Ž1 , π‘Ž2 , … … . , π‘Žπ‘› be fixed real number such that 𝑓(π‘₯) = (π‘₯ − π‘Ž1 )(π‘₯ − π‘Ž2 ) … … . (π‘₯ − π‘Žπ‘› )
what is lim 𝑓(π‘₯)? for π‘Ž ≠ π‘Ž1 , π‘Ž2 , … … π‘Žπ‘› compute lim 𝑓(π‘₯).
[NCERT]
π‘₯ + 1,
5. Let𝑓(π‘₯) = {
π‘₯ − 1,
π‘₯ + 5,
6. Let 𝑓(π‘₯) = {
π‘₯ − 4,
if π‘₯
if π‘₯
if π‘₯
if π‘₯
π‘₯→π‘Ž1
1
π‘₯→0
12. Find lim+ π‘₯−1
π‘₯→1
13. Evaluate the following one sided limits.
π‘₯−3
π‘₯−3
i) lim+ π‘₯ 2 −4
ii) lim− π‘₯ 2 −4
π‘₯→2
2
π‘₯→2
(iii) lim+
π‘₯→0
1
3π‘₯
(iv) lim +
π‘₯→−8
BY PROF. SUBHASH SONAWANE| SONAWANE ACADEMY-7722003388
2π‘₯
π‘₯+8
LIMITS
2
v) lim+
1
π‘₯5
π‘₯→0
ix) lim +
π‘₯→−2
vi) lim+ tan π‘₯
vii) lim + sec π‘₯
x) lim−(2 − cot π‘₯)
(xi) lim−1 + π‘π‘œπ‘ π‘’π‘ π‘₯
πœ‹
2
π‘₯→
π‘₯ 2 −1
2π‘₯+4
π‘₯→−
π‘₯→0
viii) lim−
π‘₯→0
πœ‹
2
π‘₯ 2 −3π‘₯+2
π‘₯ 3 −2π‘₯ 2
π‘₯→0
JEE MAIN LEVEL
1
14. Show that lim 𝑒 −π‘₯ does not exist
15. Find
π‘₯→0
(i) lim [π‘₯]
(iii) lim [π‘₯]
(ii) lim5 [π‘₯]
π‘₯→2
π‘₯→1
π‘₯→
2
16. Prove that lim+[π‘₯] = [π‘Ž] for all π‘Ž ∈ 𝑅. Also prove that lim− [π‘₯] = 0
π‘₯→π‘Ž
17. Show that lim−
18. Find
π‘₯
[π‘₯]
π‘₯→2
π‘₯
lim
it is
π‘₯→3+ [π‘₯]
π‘₯→−
π‘₯→2
equal to lim−
19. Find lim5[π‘₯]
π‘₯→1
π‘₯
≠ lim+ [π‘₯]
π‘₯→3
π‘₯
[π‘₯]
2
π‘₯ − [π‘₯], π‘₯ < 2
20. Evaluate lim 𝑓(π‘₯) (if it exists), where 𝑓(π‘₯) = { 4,
π‘₯=2
π‘₯→2
3π‘₯ − 5, π‘₯ > 2
1
21. Show that lim sin π‘₯ does not exist.
π‘₯→0
π‘˜ cos π‘₯
,
πœ‹−2π‘₯
22. Let 𝑓(π‘₯) = {
3,
where π‘₯ ≠
where π‘₯ ≠
πœ‹
2
πœ‹
2
πœ‹
2
and if limπœ‹ 𝑓(π‘₯) = 𝑓 ( ), find the value of k.
π‘₯→
2
Answer:
2) π‘˜ = 5
7) 4
8) 3, does not exist
9) does not exist
11) 0, (π‘Ž − π‘Ž1 )(π‘Ž − π‘Ž2 ) … … . . (π‘Ž − π‘Ž1 )
2) ∞
13)
i)−∞ (ii) ∞
iv)−∞ (v) ∞
vi) ∞ vii)−∞ viii)−∞
ix) ∞ x) ∞ xi)−∞
15)
i) does not exist
(ii)2
(iii)does not exist
18)1, No
19) −3
20) 1
22) π‘˜ = 6
3
BY PROF. SUBHASH SONAWANE| SONAWANE ACADEMY-7722003388
10) does not exist
(iii) ∞
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