Check sibling questions

For limits,Β 

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

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

We use limit formula to solve it.

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We have provided all formulas of limits like

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Limits of Trigonometry Functions

Limits of Trigonometry Functions.jpg

Limits of Log and Exponential Functions

Limits Formula Sheet - Part 2

Limits of the form 1 ∞ and x^n formula

Limits Formula Sheet - Part 3

Checking if Limit Exists

Limits Formula Sheet - Part 4

L'hospital’s rule

Limits Formula Sheet - Part 5

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.