Limits - Limit exists

Chapter 12 Class 11 Limits and Derivatives
Concept wise

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

## 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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