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For each t $\epsilon$ R, let [t] be the greatest integer less than or equal to t. Then

$\lim_{x\rightarrow 0}x\left ( \left [1/x\right ]\left +[2/x \right ]\left+ .....+[15/x \right ] \right )$

Option 1)
does not exist (in R).

Option 2)
is equal to 0.

Option 3)
is equal to 15

Option 4)
is equal to 120

Answers (2)

best_answer

As we learnt in

The SANDWICH THEOREM -

$If \:\:f(x)\leq g(x)\leq h(x)\:\:for\:every\:(x)\:in\:the\:deleted\:neighbourhood\:of\:(a).$

$and\:\:\:\lim_{x\rightarrow a}\:\:f(x)=\lim_{x\rightarrow a}\:\:h(x)=l$

$Then\;\:\lim_{x\rightarrow a}\:g(x)\:\:is\:also\:equal\:to\:l$

- wherein

$Where\:\:\:x\epsilon \:(a-\delta ,\:a+\delta )\:\:and\:\:\delta\: \:is\:very\:small.$

$x-1 < \left [ x \right ]\leq x$

$1/x-1 < \left [ 1/x \right ]\leq 1/x$

$2/x-1 < \left [ 2/x \right ]\leq 2/x$

.

.

.

.

$15/x-1 < \left [ 15/x \right ]\leq 15/x$

ADD ALL ;

$120/x- 15< f(n)\leq 120/x$

$120-15x/x< f(n)\leq 120/x$

$\lim_{x\rightarrow 0^{+}} (\frac{120-15x}{x})< \lim_{x\rightarrow 0^{+}}f(x)\leq \lim_{x\rightarrow 0^{+} } 120/x$

$\lim_{x\rightarrow 0^{+}} 120 -15x< \lim_{x\rightarrow 0^{+}}f(x)\leq 120$

Thus $\lim_{x\rightarrow 0}f(x)=120$

Option 1)

does not exist (in R).

Option 2)

is equal to 0.

Option 3)

is equal to 15

Option 4)

is equal to 120

Posted by

Himanshu

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