Question

If $${x_1} = 3$$  and $${x_{n + 1}} = \sqrt {2 + {x_n}} ,\,n \geqslant 1,$$     then $$\mathop {\lim }\limits_{n \to \infty } {x_n}$$  is equal to :

A. $$ - 1$$
B. $$2$$  
C. $$\sqrt 5 $$
D. $$3$$
Answer :   $$2$$
Solution :
$$\eqalign{ & {x_{n + 1}} = \sqrt {2 + {x_n}} \cr & \Rightarrow \lim {x_{n + 1}} = \sqrt {2 + \lim \,{x_n}} \cr & \Rightarrow t = \sqrt {2 + t} \,\,\,\,\,\left[ {\because \,\,\lim \,{x_{n + 1}} = \lim \,{x_n}} \right] \cr & \Rightarrow {t^2} - t - 2 = 0 \cr & \Rightarrow \left( {t - 2} \right)\left( {t + 1} \right) = 0 \cr & \Rightarrow t = 2 \cr} $$

Releted MCQ Question on
Calculus >> Limits

Releted Question 1

lf $$f\left( x \right) = \sqrt {\frac{{x - \sin \,x}}{{x + {{\cos }^2}x}}} ,$$     then $$\mathop {\lim }\limits_{x\, \to \,\infty } f\left( x \right)$$    is-

A. $$0$$
B. $$\infty $$
C. $$1$$
D. none of these
Releted Question 2

If $$G\left( x \right) = - \sqrt {25 - {x^2}} $$     then $$\mathop {\lim }\limits_{x\, \to \,{\text{I}}} \frac{{G\left( x \right) - G\left( I \right)}}{{x - 1}}$$     has the value-

A. $$\frac{1}{{24}}$$
B. $$\frac{1}{{5}}$$
C. $$ - \sqrt {24} $$
D. none of these
Releted Question 3

$$\mathop {\lim }\limits_{n\, \to \,\infty } \left\{ {\frac{1}{{1 - {n^2}}} + \frac{2}{{1 - {n^2}}} + ..... + \frac{n}{{1 - {n^2}}}} \right\}$$        is equal to-

A. $$0$$
B. $$ - \frac{1}{2}$$
C. $$ \frac{1}{2}$$
D. none of these
Releted Question 4

If $$\eqalign{ & f\left( x \right) = \frac{{\sin \left[ x \right]}}{{\left[ x \right]}},\,\,\left[ x \right] \ne 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ x \right] = 0 \cr} $$
Where \[\left[ x \right]\] denotes the greatest integer less than or equal to $$x.$$ then $$\mathop {\lim }\limits_{x\, \to \,0} f\left( x \right)$$   equals

A. $$1$$
B. $$0$$
C. $$ - 1$$
D. none of these

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Limits


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