Limits of Trigonometry Functions
limβ¬(π₯ β 0)β‘sinβ‘π₯ =0
limβ¬(π₯ β 0)β‘cosβ‘π₯ =1
limβ¬(π₯ β 0)β‘γsinβ‘π₯/π₯γ=1
limβ¬(π₯ β 0)β‘γtanβ‘π₯/π₯γ=1
limβ¬(π₯ β 0)β‘γ(1 β cosβ‘π₯)/π₯γ=0
limβ¬(π₯ β 0)β‘γsin^(β1)β‘π₯/π₯γ=1
limβ¬(π₯ β 0)β‘γtan^(β1)β‘π₯/π₯γ=1
Limits of Log and Exponential Functions
limβ¬(π₯ β 0)β‘γπ^π₯ γ=1
limβ¬(π₯ β 0)β‘γ(π^π₯ β 1)/π₯γ=1
limβ¬(π₯ β 0)β‘γ(π^π₯ β 1)/π₯γ= log_πβ‘π
limβ¬(π₯ β 0)β‘γγlog γβ‘γ(1 + π₯)γ/π₯γ=1
limβ¬(π₯ β β)β‘γ(1+1/π₯)^π₯ γ=π
limβ¬(π₯ β 0)β‘γ(1+π₯)^(1/π₯) γ=π
limβ¬(π₯ β β)β‘γ(1+π/π₯)^π₯ γ=π^π
Limits of the form π^β
limβ¬(π₯ β 0)β‘γ(1+π₯)^(1/π₯) γ=π
limβ¬(π₯ β β)β‘γ(1+1/π₯)^π₯ γ=π
limβ¬(π₯ β β)β‘γ(1+π/π₯)^π₯ γ=π^π
x^n Formula
limβ¬(π₯ β π)β‘γ((π₯^π β π^π))/(π₯ β π)γ = π(π)^(π β 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.limβ¬(γxβπγ^β ) f(x) = limβ¬(γxβπγ^+ )f(x) = f(a)
L'hospitalβs rule
If limβ¬(π₯βπ)β‘γ(π(π₯))/(π(π₯))γ gives 0/0 form
where
π(π) = 0
π(π) = 0
Then,
limβ¬(π₯βπ)β‘γ(π(π₯))/(π(π₯))γ = (π^β² (π))/(π^β² (π))
For example
limβ¬(π₯βπ)β‘γ(π₯^π β π^π)/(π₯ β π)γ
limβ¬(π₯βπ)β‘γπ₯^π βπ^π γ=0
limβ¬(π₯βπ)β‘γ(π₯βπ)γ=0
Hence it is a 0/0 form
β΄ limβ¬(π₯ β π)β‘γ((π₯^π β π^π))/(π₯ β π)γ = (π₯^π β π^π )^β²/(π₯ β π)^β²
= limβ¬(π₯ β π)β‘γ(ππ₯^(π β 1))/1γ
= ππ^(π β 1)
Made by
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.