Limx #8594;01+x+x2 1x When x = 0, the expression 1+x+x2 1x takes the form 00. Rationalising the numerator: limx #8594;01+x+x2 11+x+x2+1x1+x+x2+1=limx #8 ...

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Standard XII

Mathematics

Inequalities Involving Mathematical Means

lim x → 01+x+...

Question

$\lim_{x \to 0} \frac{\sqrt{1 + x + x^{2}} - 1}{x}$

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Solution

$\lim_{x \to 0} [\frac{\sqrt{1 + x + x^{2}} - 1}{x}]$

When x = 0, the expression $\frac{\sqrt{1 + x + x^{2}} - 1}{x}$ takes the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{1 + x + x^{2}} - 1) (\sqrt{1 + x + x^{2}} + 1)}{x (\sqrt{1 + x + x^{2}} + 1)}] = \lim_{x \to 0} [\frac{1 + x + x^{2} - 1}{x (\sqrt{1 + x + x^{2}} + 1)}] = \lim_{x \to 0} [\frac{x (1 + x)}{x (\sqrt{1 + x + x^{2}} + 1)}] = \frac{1 + 0}{\sqrt{1 + 0 + 0} + 1} = \frac{1}{2}$

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Similar questions

\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}

{lim}_{x \to 0} \frac{(1 + x)^{\frac{1}{x}} + e (x - 1)}{s i n^{- 1} x} =

(i) Find the derivative of $\sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4)}}$ + $\frac{s i n x}{1 + t a n x}$

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\lim_{x \to 0} \frac{\sqrt{1 + x + x^{2}} - \sqrt{x + 1}}{2 x^{2}}

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is equal to

(a)

\frac{1}{2}

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limx→0 1+x+x2-1x

limx→01+x+x2-1x When x = 0, the expression 1+x+x2-1x takes the form 00. Rationalising the numerator: limx→01+x+x2-11+x+x2+1x1+x+x2+1=limx→01+x+x2-1x1+x+x2+1=limx→0x1+xx1+x+x2+1=1+01+0+0+1=12

$\lim_{x \to 0} \frac{\sqrt{1 + x + x^{2}} - 1}{x}$