Limits

Calculus Level 3

lim ⁡ y → 1 y − y y 1 − y + ln ⁡ y = ? \large \lim_{y\to 1} \dfrac{ y - y^y}{1 - y + \ln y} = \, ?


The answer is 2.

Aakhyat Singh by Aakhyat Singh , 20, India

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1 solution

L = lim ⁡ y → 1 y − y y 1 − y + ln ⁡ y A 0/0 cases, L’H o ˆ pital’s rule applies. = lim ⁡ y → 1 1 − ( ln ⁡ y + 1 ) y y − 1 + 1 y A 0/0 cases again = lim ⁡ y → 1 − 1 y ⋅ y y − ( ln ⁡ y + 1 ) 2 y y − 1 y 2 Differentiate up and down w.r.t. y = 2 \begin{aligned} L & = \lim_{y \to 1} \frac {y-y^y}{1-y+\ln y} & \small \color{#3D99F6} \text{A 0/0 cases, L'Hôpital's rule applies.} \\ & = \lim_{y \to 1} \frac {1-(\ln y + 1)y^y}{-1+\frac 1y} & \small \color{#3D99F6} \text{A 0/0 cases again} \\ & = \lim_{y \to 1} \frac {-\frac 1y \cdot y^y-(\ln y + 1)^2y^y}{-\frac 1{y^2}} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }y \\ & = \boxed{2} \end{aligned}

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