Aspire Faculty ID #15195 · Topic: JAMIA MCA 2017 · Just now
JAMIA MCA 2017

$\displaystyle \lim_{x\to\infty}\Big(x-\sqrt{x^{2}+x}\,\Big)$ is equal to …

Solution

$\sqrt{x^{2}+x}=x\sqrt{1+1/x}=x\left(1+\tfrac{1}{2x}+o(1/x)\right)=x+\tfrac12+o(1)$. Hence limit $=x-(x+\tfrac12)= -\tfrac12$.

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