Question

Let f(x) be defined as follows

f(x) = $\lim_{n \rightarrow \infty}\left( \frac{n^{2} - n + 1}{n^{2} - n - 1} \right)^{n(n - 1)} =$

If f(x) is continuous at x = 0, then (a, b) =

• A. $e^{2}$
• B. $e^{- 1}$
• C. (e, e)
• D. (e^–1, e^–1)

Answer 1

Answer: (B)

$e^{- 1}$

Answer 2

We apply check for continuity at x = 0

LHL = $\lim _ { x \rightarrow 0 ^ { - } }$f(x) = (0 – h) =

(cos h + sin h)^{–cosec h} (1^∞ form)

= exp { (cos h + sin h – 1) × – cosec h}

= exp $\left\{ \lim _ { \mathrm { h } \rightarrow 0 } \left( - 2 \sin ^ { 2 } \frac { \mathrm {~h} } { 2 } + 2 \sin \frac { \mathrm {~h} } { 2 } \cos \frac { \mathrm {~h} } { 2 } \right) \times \frac { - 1 } { 2 \sin \frac { \mathrm {~h} } { 2 } \cos \frac { \mathrm {~h} } { 2 } } \right\}$

= exp= e^–1

RHL = $\lim _ { x \rightarrow 0 ^ { + } }$ f(x) = $\lim _ { h \rightarrow 0 }$ f(0 + h)

= $\frac { \mathrm { e } ^ { 1 / \mathrm { h } } + \mathrm { e } ^ { 2 / \mathrm { h } } + \mathrm { e } ^ { 3 / \mathrm { h } } } { \mathrm { ae } ^ { 2 / \mathrm { h } } + \mathrm { be } ^ { 3 / \mathrm { h } } }$

=

∴ For continuity at x = 0,

e^–1 = a = b^–1 ⇒ a = $\frac { 1 } { \mathrm { e } }$ , b = e.