Question

Lim x->0 V7+x-V7-x/5x

Answer 1

lim ( V7+x-V7-x)(V7+x + V7-x) /5x*(V7+x + V7-x)=lim 7+x-(7-x) /5x*(V7+x +V7-x)=lim 2x/5x*(V7+x+V7-x)=lim2/5*(V7+x+V7-x)=2/5*2V7=1/5V7

Answer 2

To evaluate the limit as x approaches 0 of the expression (V(7 + x) - V(7 - x))/(5x), where V denotes the square root function, we can simplify it using algebraic manipulations.

Let's start by expressing the square roots as exponents:

(V(7 + x) - V(7 - x))/(5x) = ((7 + x)^(1/2) - (7 - x)^(1/2))/(5x)

To eliminate the square roots, we can multiply the numerator and denominator by the conjugate of the numerator:

((7 + x)^(1/2) - (7 - x)^(1/2))/(5x) * ((7 + x)^(1/2) + (7 - x)^(1/2))/((7 + x)^(1/2) + (7 - x)^(1/2))

Expanding the numerator and denominator:

= [(7 + x) - (7 - x)] / [5x * ((7 + x)^(1/2) + (7 - x)^(1/2))]

= [x + x] / [5x * ((7 + x)^(1/2) + (7 - x)^(1/2))]

= [2x] / [5x * ((7 + x)^(1/2) + (7 - x)^(1/2))]

The x term cancels out:

= 2 / [5 * ((7 + x)^(1/2) + (7 - x)^(1/2))]

Now, let's evaluate the limit as x approaches 0:

lim(x->0) 2 / [5 * ((7 + x)^(1/2) + (7 - x)^(1/2))]

Plugging in x = 0:

= 2 / [5 * ((7 + 0)^(1/2) + (7 - 0)^(1/2))]

= 2 / [5 * (7^(1/2) + 7^(1/2))]

= 2 / [5 * (2 * 7^(1/2))]

= 2 / [10 * 7^(1/2)]

Simplifying:

= 1 / (5 * 7^(1/2))

= 1 / (5 * √7)

Therefore, the limit as x approaches 0 of (V(7 + x) - V(7 - x))/(5x) is 1 / (5 * √7).