Question

let f(x)=x^(2)e^(x^(2)) the value of (1)/(4)lim_(x rarr-1) f(x) ^(1/x) is

Answer 1

Step-by-step explanation:

We have,

$f(x) = {x}^{2} {e}^{ {x}^{2} }$

Now,

$\frac{1}{4} \lim_{x \rarr 1} [ {x}^{2} {e}^{ {x}^{2} } ] ^{ \frac{1}{x} }$

$= \frac{1}{4} .e$

Answer 2

To find the value of (1/4)lim_(x→-1) f(x)^(1/x), we first need to evaluate the limit of f(x)^(1/x) as x approaches -1.

Using the properties of limits and exponential functions, we can rewrite the expression as:

(1/4)lim_(x→-1) f(x)^(1/x) = (1/4) lim_(x→-1) e^(ln[f(x)^(1/x)])

= (1/4) lim_(x→-1) e^(ln[f(x)]/x)

= (1/4) e^lim_(x→-1) ln[f(x)]/x

To evaluate the limit lim_(x→-1) ln[f(x)]/x, we can use L'Hopital's rule, which states that if we have an indeterminate form of the type 0/0 or ∞/∞, then we can take the derivative of the numerator and denominator and evaluate the limit again.

So, applying L'Hopital's rule, we get:

lim_(x→-1) ln[f(x)]/x = lim_(x→-1) [2xe^(x^2) + e^(x^2)(2x^3)]/[x(x^2 + 1)e^(x^2)]

= lim_(x→-1) [2x + 2x^3]/[x(x^2 + 1)]

= lim_(x→-1) [2 + 6x^2]/[x^2 + 1]

= -4

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