Limits

Question

What is $\lim_{ x \to 0} (1-x)^{\frac{1}{x}}$ ? Also please explain the result?

Answer 1

$\lim_{ x \to 0} (1-x)^{\frac{1}{x}}$

$=\lim_{x \to 0} e^{\ln (1-x)^{\frac{1}{x}}}$

$=\lim_{x \to 0} e^{\frac {\ln  (1-x)} {x} }$

$= e^ {\lim_{x \to 0} \frac {\ln  (1-x)} {x} } \because e \text{ is a constant}$

This limit is $\frac{0}{0}$ form and we can apply L'Hôpital's rule which gives limit

$=e^{\frac{1}{-1}} = e^{-1}.$

Comments

Sir can you please explain why lim_y->infinity(1-1/y)^y=lim _y->infinity(1+1/y)^-y

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I don't have a proof for it, so I changed to another proof.

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@arjun I am getting  e^1 no negation.after l hospital it will be e^ limx->0 (1/1-x) ?

Answer 2

You can apply L hospital rule and solve .

Comments

how?

Answer 3

this is one of the best solution for this type of question when by putting limit of x in given question it forms 1^infinity form then we apply formula given below

such type of question is solved by  e^{lim x->0(f(x)-1)g(x)} 

in this question g(x)=1/x and f(x)=1-x   so , according to above formula e^{lim x->0 (1-x-1)*1/x} =e^-1

Comments

can you please share some source for this formula?

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How did you reach this formula?

Answer 4

Genral formua to solve this kind of question

lim f(x)^g(x) such that f(x)->1 and g(x)->infinite then the ans can directly be calculated by  the formula

   =e^{(f(x)-1)*g(x)}

in this case ,the ans is  e^{(1-x-1)*(1/x)}=e^-1

This formula can be applied anywhere where the form of ques is (1)^infinite

Comments

Solve and not use short cuts for GATE.

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....k sir .....i can solve this problem by using expansion or l h rule but expansion is little lengthy so i use to solve  by this readymade  formula .but now i m going to solve these ques without using these formulas ....