lim _{x rightarrow infty}left[sqrt{x^2+2 x-1} | vedclass

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Question

(D) લક્ષ $\lim _{x \rightarrow \infty}\left[\sqrt{x^2+2 x-1}-x\right]$ ની ગણતરી કરવા માટે,આપણે પદને અનુબદ્ધ $\sqrt{x^2+2 x-1}+x$ વડે ગુણી અને ભાગીશું:

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$1$

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(D) લક્ષ $\lim _{x \rightarrow \infty}\left[\sqrt{x^2+2 x-1}-x\right]$ ની ગણતરી કરવા માટે,આપણે પદને અનુબદ્ધ $\sqrt{x^2+2 x-1}+x$ વડે ગુણી અને ભાગીશું: $\lim _{x \rightarrow \infty}\left[\frac{(\sqrt{x^2+2 x-1}-x)(\sqrt{x^2+2 x-1}+x)}{\sqrt{x^2+2 x-1}+x}\right]$ $= \lim _{x \rightarrow \infty}\left[\frac{x^2+2 x-1-x^2}{\sqrt{x^2+2 x-1}+x}\right]$ $= \lim _{x \rightarrow \infty}\left[\frac{2x-1}{\sqrt{x^2(1+\frac{2}{x}-\frac{1}{x^2})}+x}\right]$ $= \lim _{x \rightarrow \infty}\left[\frac{x(2-\frac{1}{x})}{x(\sqrt{1+\frac{2}{x}-\frac{1}{x^2}}+1)}\right]$ $= \frac{2-0}{\sqrt{1+0-0}+1} = \frac{2}{2} = 1$

$\lim _{x \rightarrow \infty}\left[\sqrt{x^2+2 x-1}-x\right]$ ની કિંમત શોધો :

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લક્ષ શોધો: $\mathop {\lim }\limits_{x \to 1} (x^3 - x^2 + 1)$

જો $\alpha=\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{1-\cos x}$ અને $\beta=\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{\sqrt{1+x^2}-\sqrt{1-x^2}}$ હોય,તો

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