The Limiting limit

Calculus Level pending

Answer the question number 4. Try not to use L'Hôpital's Rule P.S. I have written a doubt in solution section please help me with it if you can.

Moderator's edit :

Find the value of α \alpha for wihch the functiion f ( x ) f(x) is defined as

f ( x ) = { α sin π 2 ( x + 1 ) , x 0 tan x sin x x 3 , x > 0 f(x) = \begin{cases} \alpha \sin \dfrac{\pi}2 (x+1) , \qquad x \leq 0 \\ \dfrac{\tan x - \sin x}{x^3} , \qquad x> 0 \end{cases}

and is continuous at x = 0 x=0 .

1.5 1 0 0.5

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2 solutions

By splitting u mean to distribute tanx and sinx over the denominator x 3 x^3 separately as two fractions?

Yes like tanx/x^3 -sinx/x^3 = 1/x^2 - 1/x^2 = 0

Somesh Patil - 5 years, 3 months ago
Somesh Patil
Mar 10, 2016

THIS IS NOT A SOLUTION BUT A DOUBT: Why can i not split the trigonometric function in numerator and get zero. Because by algebra of limits Because by algebra of limits (property one) it should work. Though one can get answer to this by L'Hôpital's Rule or the writing tanx as s i n x c o s x \frac{sinx}{cosx} then taking LCM and solve.

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