+1 vote
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What should be the value of a,b and c such that the function defined below is continuous at x=0 ?

$f\left ( x \right )=\begin{Bmatrix} \left ( 1+ax \right )^{\frac{1}{x}} & x<0 & \\ b & x=0& \\ \frac{(x+c)^{\frac{1}{3}}-1}{x} &x>0 & \end{Bmatrix}$

asked in Calculus | 183 views
a = b = infinity, if c = any finite value other than 1??

If c = 1, then , b = 1/3 , a = ln (1/3).

??
Yes right.Provide the solution by considering second case.

See.. Since it is continuous at x = 0, so f(0-) = f(0+) = f(0). ------- (X)

Here f(0-) = Lim(x->0) { (1+ax)1/x } = ea .

f(0)  = b

and  f(0+) =Lim ( x->0)  {(x + c)1/3 - 1}/ (x)

Now if c = 1 then,

f(0+) =Lim ( x->0) {(x + 1)1/3 - 1} /(x)  => 0/0

using L hospital rule, we can solve above limit = 1/3 .

Hence from eqn (X) , ea = b = 1/3.

a = ln (1/3)

b = 1/3

c = 1.

Is it okay now  ?

yes.

1 Answer

0 votes
$a=$any value

$b=1$

$c$=$x^{3}+2x+3x^{2}+1$

Then x will be continuous at x=1

But continuous in x=0 not possible here
answered by Veteran (58.2k points)
No .Actually we have given that  the above function is continuous at x=0 so  find the value of a,b,c.

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