Solution for the question - the value of f (0) so that f(x) = (1+5x)^(1//x) is continuous at x = 0 is e^(7) '/> e^(5) '/> theta^(3) '/> e^(2) '/>

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Question

The value of f (0) so that f(x) = $open parentheses 1 plus 5 x close parentheses to the power of 1 divided by x end exponent$ is continuous at x = 0 is

$e^{2}$
$e^{3}$
$e^{5}$
$e^{7}$

hint Hint:

We are given a function. We have to find its value at x = 0. To find the value of function at x = 0 we have to substitute 0 in place of x. We will use the formula of limit.

The correct answer is: $e to the power of 5 end exponent$

The given function is $f left parenthesis x right parenthesis space equals left parenthesis space 1 space plus space 5 x right parenthesis to the power of bevelled 1 over x end exponent$
The condition for given function to be continuous at any point is
$L e t space " a " space b e space a n y space p o i n t space f left parenthesis a right parenthesis space equals limit as x rightwards arrow a of space f left parenthesis x right parenthesis$
We will use this condition to solve the question.
We will substitute a = 0
$f left parenthesis 0 right parenthesis space equals space limit as x rightwards arrow 0 of space f left parenthesis x right parenthesis space space space space space space space space space equals space limit as x rightwards arrow 0 of space left parenthesis space 1 space plus space 5 x right parenthesis to the power of bevelled 1 over x end exponent T h e r e space i s space a space f o r m u l a space limit as x rightwards arrow 0 of left parenthesis space 1 space plus space k x right parenthesis to the power of bevelled 1 over x end exponent space equals space e to the power of k end exponent space space space space space space space... left parenthesis space k space i s space a n y space c o n s tan t right parenthesis S o comma space w e space g e t f left parenthesis 0 right parenthesis space equals space e to the power of 5$
So, the value of the given function at x = 0 is e⁵.

We have to be careful that denominator is not zero. We have to readjust the function in a way we will not get zero at denominator for the function to be continuous at that point. And, we have to know different formulas of limit.

If $blank f open parentheses x close parentheses equals fraction numerator e to the power of fraction numerator 1 over denominator x to the power of 2 end exponent end fraction end exponent over denominator e to the power of fraction numerator 1 over denominator x to the power of 2 end exponent end fraction end exponent minus 1 end fraction comma x not equal to 0 comma f open parentheses 0 close parentheses equals 1 blank$ then $f blank$ at $blank x equals 0 blank$ is

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$Error converting from MathML to accessible text.$ equal to

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The value of f (0) so that f(x) = is continuous at x = 0 is

We are given a function. We have to find its value at x = 0. To find the value of function at x = 0 we have to substitute 0 in place of x. We will use the formula of limit.

The correct answer is:

The given function is The condition for given function to be continuous at any point isWe will use this condition to solve the question.We will substitute a = 0So, the value of the given function at x = 0 is e5.

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