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$\lim_{x\rightarrow \frac{\pi }{4}}\frac{\cot ^{3}x-\tan x}{\cos \left ( x+\frac{\pi }{4} \right )}$ is :
Option 1)
$4\sqrt{2}$

Option 2)
$8$
Option 3)
$4$
Option 4)
$8\sqrt{2}$

Answers (1)

best_answer

Evalution of Trigonometric limit -

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

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

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

$Then\:it\:comes$

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

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

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

-

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

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

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

-

Using LH Rule

$\lim_{x\rightarrow \frac{\pi}{4}}\frac{3\cot ^{2}x(-\csc ^{2}x-\sec ^{2}x)}{-\sin (x+\frac{\pi}{4})}=8$

Option 1)

$4\sqrt{2}$

Option 2)
$8$
Option 3)

$4$

Option 4)

$8\sqrt{2}$

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