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Evaluate $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\cot{\Big(\dfrac{\pi}{4}+h\Big)}-1}{h}}$

In calculus, there is no trigonometric limit formula in terms of cot function but there is a formula in terms of tan function. So, try to express the cot of compound angle function in the form of tan function.

Convert the Cot function in terms of tan function

The cot function can be written in the form of tan function by the reciprocal identity of tan function and then simplify the expression.

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{1}{\tan{\Big(\dfrac{\pi}{4}+h\Big)}}-1}{h}}$

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{1-\tan{\Big(\dfrac{\pi}{4}+h\Big)}}{\tan{\Big(\dfrac{\pi}{4}+h\Big)}}}{h}}$

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1-\tan{\Big(\dfrac{\pi}{4}+h\Big)}}{h\tan{\Big(\dfrac{\pi}{4}+h\Big)}}} $

Expand tan of sum of angles

The tan of compound angle can be expanded as per the compound angle trigonometric identity.

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{1-\tan{h}-(1+\tan{h})}{1-\tan{h}}}{\dfrac{h(1+\tan{h})}{1-\tan{h}}}}$

Now, simplify the function in fraction form to evaluate the limit of this function as $h$ tends to zero.

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{1-\tan{h}-1-\tan{h}}{1-\tan{h}}}{\dfrac{h(1+\tan{h})}{1-\tan{h}}}}$

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{1-1-2\tan{h}}{1-\tan{h}}}{\dfrac{h(1+\tan{h})}{1-\tan{h}}}}$

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \require{cancel} \dfrac{\dfrac{\cancel{1}-\cancel{1}-2\tan{h}}{1-\tan{h}}}{\dfrac{h(1+\tan{h})}{1-\tan{h}}}}$

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{-2\tan{h}}{1-\tan{h}} \times \dfrac{1-\tan{h}}{h(1+\tan{h})}}$

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \require{cancel} \dfrac{-2\tan{h}}{\cancel{1-\tan{h}}} \times \dfrac{\cancel{1-\tan{h}}}{h(1+\tan{h})}}$

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{-2\tan{h}}{h(1+\tan{h})}}$

$= \,\,\,$ $\displaystyle -2\large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan{h}}{h(1+\tan{h})}}$

Factorize the function

Separate the function as two factors by factorisation because the part of the function is in the form of a trigonometric limit formula.

$= \,\,\,$ $\displaystyle -2\large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{\tan{h}}{h} \times \dfrac{1}{(1+\tan{h})} \Bigg] }$

Apply Product Rule of Limits

The limit condition belongs to both functions as per product rule of limits. So, it can written to both factors separately.

$= \,\,\,$ $\displaystyle -2 \Bigg[\large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan{h}}{h}} \times \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1}{1+\tan{h}} \Bigg] }$

Find the Limit of the function

As per limit of tanx/x as x approaches 0 formula, the value of the first factor is one. Substitute, $h$ is equal to $0$ in the second function to find the limit of the whole function in calculus mathematically.

$= \,\,\,$ $-2 \Bigg[1 \times \dfrac{1}{1+\tan{(0)}} \Bigg] $

$= \,\,\,$ $-2 \Bigg[1 \times \dfrac{1}{1+0} \Bigg]$

$= \,\,\,$ $-2 \Bigg[1 \times \dfrac{1}{1} \Bigg]$

$= \,\,\,$ $-2{(1 \times 1)}$

$= \,\,\,$ $-2 \times 1$

$= \,\,\,$ $-2$

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