Show that lim x -x2-x-1-x-lim x -x2-1-x

RHS lim_x→∞ ( √(x^2+1) x ) By Rationalising, =lim_x→∞ ( √(x^2+1) x ) ( √(x^2+1)+x )/ ( √(x^2+1)+x ) =lim_x→∞x^2+1 x^2/x√(1+1/x^2)+x =lim_x→∞1/x√(1+1/x^2)+x ...

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Byju's Answer

Standard XII

Mathematics

Rationalization Method to Remove Indeterminate Form

Show that lim...

Question

Show that ${lim}_{x \to \infty} (\sqrt{x^{2} + x + 1 - x}) \neq {lim}_{x \to \infty} (\sqrt{x^{2} + 1} - x)$

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Solution

RHS ${lim}_{x \to \infty} (\sqrt{x^{2} + 1} - x)$

By Rationalising,

$= {lim}_{x \to \infty} \frac{(\sqrt{x^{2} + 1} - x) (\sqrt{x^{2} + 1} + x)}{(\sqrt{x^{2} + 1} + x)}$

$= {lim}_{x \to \infty} \frac{x^{2} + 1 - x^{2}}{x \sqrt{1 + \frac{1}{x^{2}}} + x}$

$= {lim}_{x \to \infty} \frac{1}{x \sqrt{1 + \frac{1}{x^{2}}} + x}$

$= {lim}_{x \to \infty} \frac{1}{x} ⎛ ⎜ ⎝ \frac{1}{\sqrt{1 + \frac{1}{x^{2}}} + 1} ⎞ ⎟ ⎠ = 0$

Also,

LHS= $= {lim}_{x \to \infty} (\sqrt{x^{2} + x + 1} - x)$

$= {lim}_{x \to \infty} \frac{(\sqrt{x^{2} + x + 1} - x) (\sqrt{x^{2} + x + 1} + x)}{(\sqrt{x^{2} + x + 1} + x)}$

$= {lim}_{x \to \infty} \frac{(x^{2} + x + 1 - x^{2})}{\sqrt{x^{2} + x + 1} + x}$

$= {lim}_{x \to \infty} \frac{x (1 + \frac{1}{x})}{x (\sqrt{1 + \frac{1}{x} + \frac{1}{x^{2}}} + 1)}$

$= {lim}_{x \to \infty} \frac{1 + \frac{1}{x}}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^{2}}} + 1}$

$= \frac{1}{1 + 1} = \frac{1}{2}$

$L H S \neq R H S$

$∴ {lim}_{x \to \infty} (\sqrt{x^{2} + x + 1} - x)$ is not equal to ${lim}_{x \to \infty} (\sqrt{x^{2} + x} - x)$ .

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Similar questions

Q. Show that

\lim_{x \to \infty} (\sqrt{x^{2} + x + 1} - x) \neq \lim_{x \to \infty} (\sqrt{x^{2} + 1} - x)

Q. Show that

{lim}_{x \to 2} \frac{x^{2} - 5 x + 6}{x^{2} - x - 2} = \frac{- 1}{3}

Q. Find the limits :

i .

lim x \to 1 [\frac{x^{2} + 1}{x + 100}]

i i .

lim x \to 2 [\frac{x^{3} - 4 x^{2} + 4 x}{x^{2} - 4}]

i i i .

lim x \to 2 [\frac{x^{2} - 4}{x^{3} - 4 x^{2} + 4 x}]

i v .

lim x \to 2 [\frac{x^{3} - 2 x^{2}}{x^{2} - 5 x + 6}]

v .

lim x \to 1 [\frac{x - 2}{x^{2} - x} - \frac{1}{x^{3} - 3 x^{2} + 2 x}]

Evaluate the following one sided limits:

$(i) {lim}_{x \to 2^{+}} \frac{x - 3}{x^{2} - 4}$

$(i i) {lim}_{x \to 2^{-}} \frac{x - 3}{x^{2} - 4}$

$(i i i) {lim}_{x \to 0^{+}} \frac{1}{3 x}$

$(i v) {lim}_{x \to 8^{+}} \frac{2 x}{x + 8}$

$(v) {lim}_{x \to 0^{+}} \frac{2}{x^{\frac{1}{5}}}$

$(v i) {lim}_{x \to \frac{π^{-}}{2}} t a n x$

$(v i i) {lim}_{x \to \frac{π}{2^{+}}} s e c x$

$(v i i i) {lim}_{x \to 0^{-}} \frac{x^{2} - 3 x + 2}{x^{3} - 2 x^{2}}$

$(i x) {lim}_{x \to - 2^{+}} \frac{x^{2} - 1}{2 x + 4}$

$(x) {lim}_{x \to 0^{+}} (2 - c o t x)$

$(x i) {lim}_{x \to 0^{-}} 1 + c o s e c x$

Q. Assume that exists

{lim}_{x \to - 1} f (x)

and

\frac{x^{2} + x - 2}{x + 3} \leq \frac{f (x)}{x^{2}} \leq \frac{x^{2} + 2 x - 1}{x + 3}

holds for certain interval containing that point

x = 1,

then

{lim}_{x \to - 1} f (x)

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Show that limx→∞(√x2+x+1−x)≠limx→∞(√x2+1−x)

Show that ${lim}_{x \to \infty} (\sqrt{x^{2} + x + 1 - x}) \neq {lim}_{x \to \infty} (\sqrt{x^{2} + 1} - x)$