यदि lim _{x rightarrow 0^{+}} x^2left(frac{e^ | vedclass

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Question

(A) $k = \lim _{x \rightarrow 0^{+}} x^2 \left( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \right)$ के लिए,$t = 1/x$ लें। जैसे $x \rightarrow 0^{+}

Vedclass · Accepted Answer

$k=l=0$

Vedclass · Answer

(A) $k = \lim _{x ightarrow 0^{+}} x^2 \left( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} ight)$ के लिए,$t = 1/x$ लें। जैसे $x ightarrow 0^{+}$,$t ightarrow \infty$. $k = \lim _{t ightarrow \infty} \frac{1}{t^2} \left( \frac{e^t - e^{-t}}{e^t + e^{-t}} ight) = \lim _{t ightarrow \infty} \frac{1}{t^2} \left( \frac{1 - e^{-2t}}{1 + e^{-2t}} ight) = 0 	imes 1 = 0$. $l = \lim _{x ightarrow 0^{-}} x^2 \left( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} ight)$ के लिए,$y = 1/x$ लें। जैसे $x ightarrow 0^{-}$,$y ightarrow -\infty$. $l = \lim _{y ightarrow -\infty} \frac{1}{y^2} \left( \frac{e^y - e^{-y}}{e^y + e^{-y}} ight) = \lim _{y ightarrow -\infty} \frac{1}{y^2} \left( \frac{e^{2y} - 1}{e^{2y} + 1} ight) = 0 	imes (-1) = 0$. अतः,$k = l = 0$.

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