Check sibling questions

For limits,Β 

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

Β 

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


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 12 years. He provides courses for Maths and Science at Teachoo.