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

$\lim_{x-1+}\frac{\left ( 1-\left | x \right | +\sin\left | 1-x \right |\right )\sin\left ( \frac{\pi }{2}\left [ 1-x \right ] \right )}{\left | 1-x \right |\left | 1-x \right |}$
Option 1)

equals 1
Option 2)

equalss0
Option 3)

equals-1
Option 4)

doesnot exist

Answers (1)

best_answer

Limit of product / quotient -

Limit of product/quotient is the product/quotient of individual limits such that

$\lim_{x\rightarrow a}{\left (f(x).g(x) \right )}$

$=\lim_{x\rightarrow a}{f(x).\lim_{x\rightarrow a}g(x)$ , given that f(x) and g(x) are non-zero finite values

$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)}$ , given that f(x) and g(x) are non-zero finite values

$Also\:\lim_{x\rightarrow a}{kf(x)}$

$=k\lim_{x\rightarrow a}{f(x)}$

-

Evalution of Trigonometric limit -

$\lim_{x\rightarrow a}\:\frac{sin(x-a)}{x-a}=1$

$\lim_{x\rightarrow a}\:\frac{tan(x-a)}{x-a}=1$

$put\:\:\:\:\:x=a+h\:\:\:where\:\:h\rightarrow 0$

$Then\:it\:comes$

$\lim_{h\rightarrow 0}\:\:\frac{sinh}{h}=\lim_{h\rightarrow 0}\:\:\frac{tanh}{h}=1$

$\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{sinx}{x}=1\:\;\;and$

$\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{tanx}{x}=1$

-

$\lim_{x\rightarrow 1^{+}} \frac{\left ( 1 - \left | x \right |+ sin\left | 1 - x \right | sin\left ( \frac{\pi}{2}\left [ 1 - x \right ] \right )\right )}{\left | 1-x \right |\left [ 1-x \right ]}$

$=\lim_{x\rightarrow 1^{+}} \frac{(1-x) + sin(x-1)}{(x-1)(-1)}sin \left ( \frac{\pi}{2}(-1) \right )$

$=\lim_{x\rightarrow 1^{+}} \left ( 1 - \frac{sin(x-1)}{x-1} \right )(-1) = 0$

Option 1)

equals 1

Option 2)

equalss0
Option 3)

equals-1

Option 4)

doesnot exist

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