Question

lim x tends to alpha (x cosec alpha - alpha cosec x) /x- alpha​

Answer 1

$\large\underline{\sf{Solution-}} \:$

$\rm \: \displaystyle\lim_{x \to \alpha }\rm \frac{x \: cosec \alpha - \alpha cosecx}{x - \alpha }$

can be rewritten as

$\rm \: = \:\displaystyle\lim_{x \to \alpha }\rm \dfrac{ \dfrac{x}{sin\alpha } - \dfrac{\alpha }{sinx} }{x - \alpha }$

can be further rewritten as

$\rm \: = \:\displaystyle\lim_{x \to \alpha }\rm \dfrac{ \dfrac{xsinx - \alpha sin\alpha }{sin\alpha sinx}}{x - \alpha }$

can be further rewritten as

$\rm \: = \: \frac{1}{ {sin}^{2}\alpha } \displaystyle\lim_{x \to \alpha }\rm \frac{xsinx - \alpha sin\alpha }{x - \alpha }$

Now, to evaluate further, we use method of Substitution.

So, Substitute

$\rm \: x = \alpha + h, \: \: as \: x \to \: \alpha , \: \: so \: h \to \: 0$

So, on substituting these values, we get

$\rm \: = \: \frac{1}{ {sin}^{2}\alpha } \displaystyle\lim_{h \to 0 }\rm \frac{(\alpha + h)sin(\alpha + h) - \alpha sin\alpha }{\alpha + h - \alpha }$

$\rm \: = \: \frac{1}{ {sin}^{2}\alpha } \displaystyle\lim_{h \to 0 }\rm \frac{\alpha sin(\alpha + h) + hsin(\alpha + h) - \alpha sin\alpha }{h}$

$\rm \: = \: \frac{1}{ {sin}^{2}\alpha } \displaystyle\lim_{h \to 0 }\rm \bigg[\frac{\alpha sin(\alpha + h) - \alpha sin\alpha }{h} + \frac{hsin(\alpha + h)}{h}\bigg]$

$\rm \: = \: \frac{1}{ {sin}^{2}\alpha } \displaystyle\lim_{h \to 0 }\rm \bigg[\frac{\alpha (sin(\alpha + h) - sin\alpha) }{h} + sin(\alpha + h)\bigg]$

We know,

$\boxed{\sf{  \:\rm \: sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg] \: \: }}$

So, using this result, we get

$\rm \: = \: \frac{1}{ {sin}^{2}\alpha } \displaystyle\lim_{h \to 0 }\rm \bigg[\frac{2\alpha cos\bigg[\dfrac{\alpha + h + \alpha }{2} \bigg]sin\bigg[\dfrac{\alpha + h - \alpha }{2} \bigg]}{h}\bigg] + \frac{1}{ {sin}^{2}\alpha } \times sin\alpha\bigg]$

$\rm \: = \: \frac{1}{ {sin}^{2}\alpha } \displaystyle\lim_{h \to 0 }\rm \bigg[\frac{\alpha cos\bigg[\dfrac{2\alpha + h }{2} \bigg]sin\bigg[\dfrac{ h }{2} \bigg]}{ \dfrac{h}{2} }\bigg] + \frac{1}{ {sin}\alpha }$

$\rm \: = \: \frac{1}{ {sin}^{2}\alpha } \times \displaystyle\lim_{h \to 0}\rm \alpha cos\bigg[\dfrac{2\alpha + h}{2} \bigg] \times \displaystyle\lim_{h \to 0 }\rm \bigg[\frac{sin\bigg[\dfrac{ h }{2} \bigg]}{ \dfrac{h}{2} }\bigg] + \frac{1}{ {sin}\alpha }$

We know,

$\boxed{\sf{  \:\rm \: \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1 \: \: }}$

So, using this, we get

$\rm \: = \: \frac{\alpha cos\alpha }{ {sin}^{2}\alpha } \times 1 + \frac{1}{ {sin}\alpha }$

$\rm \: = \: \frac{\alpha cos\alpha }{ {sin}\alpha \times sin\alpha } + cosec\alpha$

$\rm \: = \:\alpha cosec\alpha \: cot\alpha \: + \: cosec\alpha$

Hence,

$\\ \rm \: \displaystyle\lim_{x \to \alpha }\rm \frac{x \: cosec \alpha - \alpha cosecx}{x - \alpha } = \alpha cosec\alpha \: cot\alpha + cosec\alpha$

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Additional Information :-

$\boxed{\sf{  \:\rm \: \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1 \: \: }}$

$\boxed{\sf{  \:\rm \: \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} = 1 \: \: }}$

$\boxed{\sf{  \:\rm \: \displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} = 1 \: \: }}$

$\boxed{\sf{  \:\rm \: \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} = 1 \: \: }}$

$\boxed{\sf{  \:\rm \: \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga \: \: }}$

Answer 2

Answer:

Value of the limit = [cosec(α) + α cosec (α) cot (α)]

Step-by-step explanation:

A very basic limits question to let you identify easily the indeterminate forms to which we can apply L'H rule and practice a bit to feel comfortable with this rule to so as to help you ahead in complex limit questions.

So,in this limit if we substitute the limiting value of x,

=> lim x → α [ x cosec(α) - α cosec(x)]/ x - α

=> [ α cosec(α) - α cosec (α)]/ α - α

=> 0/0

So,we have a 0/0 form here,so we have the option to apply the L'h rule.

So,L'h rule you can't actually evaluate the limit but it can simplify your function drastically so as to make your task very simple.

If you have a function suppose h(x) and it is expressed in some f(x)/g(x) then the h(x) can be written as f'(x)/g'(x). This rule can be applied only to f(x)/g(x) type expressed functions.

Now, let's focus on the question given to solve.

Differentiate the given function,

=> lim x → α {[d(x • cosec (α))/dx] - [d(α cosec (x)/dx]}/ [d(x - α)/dx]

=> lim x → α [cosec (α) - (- α cosec (x) cot (x)]/ 1

=> lim x → α [cosec α + α cosec (x) cot (x)]

Now,we can directly substitute the limiting value of x, since now there ain't any denominator present.

=> cosec(α) + α cosec (α) cot (α)