lim _{x rightarrow infty}left[frac{x^2+x+3}{x | vedclass

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(D) આપેલ છે,$L = \lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x$. આ $1^\infty$ સ્વરૂપ છે. સૂત્ર $\lim _{x \rightarrow \infty}[1+f(

Vedclass · Accepted Answer

$e^2$

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(D) આપેલ છે,$L = \lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x$. આ $1^\infty$ સ્વરૂપ છે. સૂત્ર $\lim _{x \rightarrow \infty}[1+f(x)]^{g(x)} = e^{\lim _{x \rightarrow \infty} f(x) \cdot g(x)}$ નો ઉપયોગ કરતા: $L = \lim _{x \rightarrow \infty}\left[1+\frac{x^2+x+3}{x^2-x+2}-1\right]^x$ $= \lim _{x \rightarrow \infty}\left[1+\frac{x^2+x+3-x^2+x-2}{x^2-x+2}\right]^x$ $= \lim _{x \rightarrow \infty}\left[1+\frac{2x+1}{x^2-x+2}\right]^x$ $= e^{\lim _{x \rightarrow \infty} \frac{2x+1}{x^2-x+2} \cdot x}$ $= e^{\lim _{x \rightarrow \infty} \frac{2x^2+x}{x^2-x+2}}$ અંશ અને છેદને $x^2$ વડે ભાગતા: $= e^{\lim _{x \rightarrow \infty} \frac{2+1/x}{1-1/x+2/x^2}} = e^{\frac{2+0}{1-0+0}} = e^2$.

$\lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x$ ની કિંમત શોધો.

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