"Let [t] denote the greatest integer $\le$ t. If for some $\lambda$ $\in$ R - {1, 0}, $$\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + \left[ x \right]}}} \right|$$ = L, then L is equal to :"

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"Here $$\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + [x]}}} \right| = L$$

Here L.H.L. $$\mathop {\lim }\limits_{h \to 0^-} \left| {{{1 + h + h} \over {\lambda + h - 1}}} \right| = \left| {{1 \over {\lambda - 1}}} \right|$$

R.H.L. = $$\mathop {\lim }\limits_{h \to 0^+} \left| {{{1 - h + h} \over {\lambda + h + 0}}} \right| = \left| {{1 \over \lambda }} \right|$$

$\because$ Limit exists. Hence L.H.L. = R.H.L.

$\Rightarrow$ $\left| {\lambda - 1} \right| = \left| \lambda \right|$

$\Rightarrow$ $\lambda = {1 \over 2}$

$\therefore$ L = ${1 \over {\left| \lambda \right|}}$ = 2"

Let [t] denote the greatest integer $\le$ t. If for some
$\lambda$ $\in$ R - {1, 0}, $$\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + \left[ x \right]}}} \right|$$ = L, then L is equal to :

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Let [t] denote the greatest integer $\le$ t. If for some $\lambda$ $\in$ R - {1, 0}, $$\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + \left[ x \right]}}} \right|$$ = L, then L is equal to :

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More Limits, Continuity and Differentiability questions

Drill 25 more like these. Every day. Free.

Let [t] denote the greatest integer $\le$ t. If for some
$\lambda$ $\in$ R - {1, 0}, $$\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + \left[ x \right]}}} \right|$$ = L, then L is equal to :