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

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Question

(A) $k = \lim _{x \rightarrow 0^{+}} x^2 \left( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \right)$ માટે,$t = 1/x$ લો. જેમ $x \rightarrow 0^{+}$,તે

Vedclass · Accepted Answer

$k=l=0$

Vedclass · Answer

(A) $k = \lim _{x ightarrow 0^{+}} x^2 \left( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} ight)$ માટે,$t = 1/x$ લો. જેમ $x ightarrow 0^{+}$,તેમ $t ightarrow \infty$. $k = \lim _{t ightarrow \infty} \frac{1}{t^2} \left( \frac{e^t - e^{-t}}{e^t + e^{-t}} ight) = \lim _{t ightarrow \infty} \frac{1}{t^2} \left( \frac{1 - e^{-2t}}{1 + e^{-2t}} ight) = 0 	imes 1 = 0$. $l = \lim _{x ightarrow 0^{-}} x^2 \left( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} ight)$ માટે,$y = 1/x$ લો. જેમ $x ightarrow 0^{-}$,તેમ $y ightarrow -\infty$. $l = \lim _{y ightarrow -\infty} \frac{1}{y^2} \left( \frac{e^y - e^{-y}}{e^y + e^{-y}} ight) = \lim _{y ightarrow -\infty} \frac{1}{y^2} \left( \frac{e^{2y} - 1}{e^{2y} + 1} ight) = 0 	imes (-1) = 0$. તેથી,$k = l = 0$.

જો $\lim _{x \rightarrow 0^{+}} x^2\left(\frac{e^{1 / x}-e^{-1 / x}}{e^{1 / x}+e^{-1 / x}}\right)=k$ અને $\lim _{x \rightarrow 0^{-}} x^2\left(\frac{e^{1 / x}-e^{-1 / x}}{e^{1 / x}+e^{-1 / x}}\right)=l$ હોય,તો નીચેનામાંથી કયું સત્ય છે?

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