Limits Formula Sheet
Last updated at July 23, 2019 by Teachoo
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
Log and Exponential Functions
Limits of the form
1
∞
and x^n formula
Checking if Limit Exists
L'hospital’s
rule
Transcript
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)
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