For limits, 

we put value and check if it is of the form 0/0, ∞/∞, 1

 

If it is of that form, we cannot find limits by putting values.

We use limit formula to solve it.

 

We have provided all formulas of limits like

 

Limits of Trigonometry Functions

Limits of Trigonometry Functions.jpg

Limits of Log and Exponential Functions

Limits of Log and Exponential Functions.jpg

Limits of the form 1 and x^n formula

Limits of the form 1 to the power infinity and xn formula.jpg

Checking if Limit Exists

Checking if Limit Exists.jpg

L'hospital’s rule

L Hospitals Rule.jpg

  1. Chapter 13 Class 11 Limits and Derivatives
  2. Concept wise

Transcript

Limits of Trigonometry Functions lim┬(π‘₯ β†’ 0)⁑sin⁑π‘₯ =0 lim┬(π‘₯ β†’ 0)⁑cos⁑π‘₯ =1 lim┬(π‘₯ β†’ 0)⁑〖sin⁑π‘₯/π‘₯γ€—=1 lim┬(π‘₯ β†’ 0)⁑〖tan⁑π‘₯/π‘₯γ€—=1 lim┬(π‘₯ β†’ 0)⁑〖(1 βˆ’ cos⁑π‘₯)/π‘₯γ€—=0 lim┬(π‘₯ β†’ 0)⁑〖sin^(βˆ’1)⁑π‘₯/π‘₯γ€—=1 lim┬(π‘₯ β†’ 0)⁑〖tan^(βˆ’1)⁑π‘₯/π‘₯γ€—=1 Limits of Log and Exponential Functions lim┬(π‘₯ β†’ 0)⁑〖𝑒^π‘₯ γ€—=1 lim┬(π‘₯ β†’ 0)⁑〖(𝑒^π‘₯ βˆ’ 1)/π‘₯γ€—=1 lim┬(π‘₯ β†’ 0)⁑〖(π‘Ž^π‘₯ βˆ’ 1)/π‘₯γ€—= log_π‘’β‘π‘Ž lim┬(π‘₯ β†’ 0)⁑〖〖log 〗⁑〖(1 + π‘₯)γ€—/π‘₯γ€—=1 lim┬(π‘₯ β†’ ∞)⁑〖(1+1/π‘₯)^π‘₯ γ€—=𝑒 lim┬(π‘₯ β†’ 0)⁑〖(1+π‘₯)^(1/π‘₯) γ€—=𝑒 lim┬(π‘₯ β†’ ∞)⁑〖(1+π‘Ž/π‘₯)^π‘₯ γ€—=𝑒^π‘Ž Limits of the form 𝟏^∞ lim┬(π‘₯ β†’ 0)⁑〖(1+π‘₯)^(1/π‘₯) γ€—=𝑒 lim┬(π‘₯ β†’ ∞)⁑〖(1+1/π‘₯)^π‘₯ γ€—=𝑒 lim┬(π‘₯ β†’ ∞)⁑〖(1+π‘Ž/π‘₯)^π‘₯ γ€—=𝑒^π‘Ž x^n Formula lim┬(π‘₯ β†’ π‘Ž)⁑〖((π‘₯^𝑛 βˆ’ π‘Ž^𝑛))/(π‘₯ βˆ’ π‘Ž)γ€— = 𝑛(π‘Ž)^(𝑛 βˆ’ 1) To check if limit exists for f(x) at x = a We check if Left Hand Limit = Right Hand Limit = f(a) i.e.lim┬(γ€–xβ†’π‘Žγ€—^βˆ’ ) f(x) = lim┬(γ€–xβ†’π‘Žγ€—^+ )f(x) = f(a) L'hospital’s rule If lim┬(π‘₯β†’π‘Ž)⁑〖(𝑓(π‘₯))/(𝑔(π‘₯))γ€— gives 0/0 form where 𝑓(π‘Ž) = 0 𝑔(π‘Ž) = 0 Then, lim┬(π‘₯β†’π‘Ž)⁑〖(𝑓(π‘₯))/(𝑔(π‘₯))γ€— = (𝑓^β€² (π‘Ž))/(𝑔^β€² (π‘Ž)) For example lim┬(π‘₯β†’π‘Ž)⁑〖(π‘₯^𝑛 βˆ’ π‘Ž^𝑛)/(π‘₯ βˆ’ π‘Ž)γ€— lim┬(π‘₯β†’π‘Ž)⁑〖π‘₯^𝑛 βˆ’π‘Ž^𝑛 γ€—=0 lim┬(π‘₯β†’π‘Ž)⁑〖(π‘₯βˆ’π‘Ž)γ€—=0 Hence it is a 0/0 form ∴ lim┬(π‘₯ β†’ π‘Ž)⁑〖((π‘₯^𝑛 βˆ’ π‘Ž^𝑛))/(π‘₯ βˆ’ π‘Ž)γ€— = (π‘₯^𝑛 βˆ’ π‘Ž^𝑛 )^β€²/(π‘₯ βˆ’ π‘Ž)^β€² = lim┬(π‘₯ β†’ π‘Ž)⁑〖(𝑛π‘₯^(𝑛 βˆ’ 1))/1γ€— = π‘›π‘Ž^(𝑛 βˆ’ 1)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.