S0NAWANE ACADEMY MATHS COACHING CHAPTER- limits JOB SHEET NO- -01 DEFINITION OF LIMIT: lim π(π₯) exists iff π₯→π lim lim π(π₯) = π(π − β) exists, π₯ → π− β→0 lim lim (2) R.H.L.=Right hand limit= π(π+ ) = π(π₯) = π(π + β) exists, π₯ → π+ β→0 lim lim (3) π(π₯). − π(π₯) = π₯→π π₯ → π+ lim lim If lim π(π₯) exists then lim π(π₯) = π(π₯) = π(π₯). π₯→π π₯→π π₯ → π− π₯ → π+ (1) L.H.L.=Left hand limit = π(π− ) = Q.1 1] Evaluate: lim π(π₯) where π(π₯) = π₯→π π₯ 2 −4 at π₯ = 2 π₯−2 |π₯−4| , π₯≠4 2] Evaluate: lim π(π₯) where π(π₯) = { π₯−4 ππ‘ π₯ = 4 π₯→π 0, π₯=4 1 + π₯2, 0 ≤ π₯ ≤ 1 3]Evaluate: lim π(π₯) where π(π₯) = { ππ‘ π₯ = 1 π₯→π 2 − π₯, π₯>1 π₯−|π₯| , π₯≠0 4]If π(π₯) = { π₯ show that lim π(π₯) does not exist. π₯→0 2, π₯=0 5π₯ − 4, 0 < π₯ ≤ 1 5]If π(π₯) = { 3 show that lim π(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 πππ π₯, π₯>0 7] Let π(π₯) = { find the value of constant k , π₯ + π, π₯<0 given that lim π(π₯) exists. π₯→0 8]Let π(π₯) be function defined by 4π₯ − 5, π₯ ≤ 2 π(π₯) = { find πΌ if lim π(π₯) exists. π₯ − πΌ, π₯>2 π₯→2 ππ₯ 2 + π, π₯ < 0 9] If π(π₯) = {ππ₯ + π, 0≤π₯≤1 3 ππ₯ + π, π₯ > 1 For what values of integers m and n does the limits lim π(π₯) and lim π(π₯) exists. π₯→0 π₯→1 [NCERT] LIMITS |π₯| + 1, π₯ < 0 π₯ = 0 for what value of a does lim π(π₯)exist ? [NCERT] 10] If π(π₯) = { 0, π₯→0 |π₯| − 1, π₯ > 0 π + ππ₯, π₯ < 1 11] Suppose π(π₯) = { 4, π₯=1 π − ππ₯, π₯ > 1 And if lim π(π₯) = π(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 π(π₯) = [π₯]= greatest intger less than or equal to , at π₯ = π , where π is an ineteger. Also show that lim π(π₯) does not exists. π₯→π 2] Prove that lim+[π₯] = [π] for all ∈ π , where [. ] denotes the greatest integer function. π₯→π π 1/π₯ −1 3] Show that lim π 1/π₯ +1 does not exist. π₯→0 4] If πis an odd function and if lim π(π₯) exists. Prove that this limit must be zero. π₯→0 5] If πis an even function then prove that lim− π(π₯) = lim+ π(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 π(π₯) may exist, where (π₯) = { . π₯ + π, π₯ > 2 π₯→2 1 3. Show that lim π₯ does not exist. π₯→0 4. Let π(π₯) be a function defined by π(π₯) = 3π₯ {|π₯||2π₯ , π₯≠0 0, π₯=0 >0 . Prove that lim π(π₯) does not exist. <0 π₯→0 >0 prove that lim π(π₯) does not exist. <0 π₯→2 4, if π₯ > 1 7. Find lim π(π₯) where π(π₯) = { π₯ + 1, if π₯ < 1 π₯→1 2π₯ + 3, π₯ ≤ 0 8. If π(π₯) = { find lim π(π₯) and lim π(π₯) [NCERT] 3(π₯ + 1), π₯ > 0 π₯→0 π₯→1 π₯ 2 − 1, π₯ ≤ 1 9. Find lim π(π₯), if (π₯) = { 2 . [NCERT] π₯→1 −π₯ − 1, π₯ > 1 |π₯| , π₯≠0 10. Evaluate lim π(π₯), where π(π₯) = { π₯ [NCERT] π₯→0 0, π₯ = 0 11. Let π1 , π2 , … … . , ππ be fixed real number such that π(π₯) = (π₯ − π1 )(π₯ − π2 ) … … . (π₯ − ππ ) what is lim π(π₯)? for π ≠ π1 , π2 , … … ππ compute lim π(π₯). [NCERT] π₯ + 1, 5. Letπ(π₯) = { π₯ − 1, π₯ + 5, 6. Let π(π₯) = { π₯ − 4, if π₯ if π₯ if π₯ if π₯ π₯→π1 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 + πππ ππ π₯ π 2 π₯→ π₯ 2 −1 2π₯+4 π₯→− π₯→0 viii) lim− π₯→0 π 2 π₯ 2 −3π₯+2 π₯ 3 −2π₯ 2 π₯→0 JEE MAIN LEVEL 1 14. Show that lim π −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] for all π ∈ π . Also prove that lim− [π₯] = 0 π₯→π 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 π(π₯) (if it exists), where π(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−2π₯ 22. Let π(π₯) = { 3, where π₯ ≠ where π₯ ≠ π 2 π 2 π 2 and if limπ π(π₯) = π ( ), 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 − π1 )(π − π2 ) … … . . (π − π1 ) 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) ∞