Question

If F: R→ R is a differentiable function and f(3) = 6, then $\lim_{x\to3} \int_{6}^{f(x)} \frac{2tdt}{(x-2)}$ is equal to

Answer 1

0

Answer 2

Let the given limit be $L$. The expression is $\lim_{x\to3} \int_{6}^{f(x)} \frac{2tdt}{(x-2)}$. The term $(x-2)$ in the denominator is constant with respect to the integration variable $t$. So we can take it out of the integral: $L = \lim_{x\to3} \frac{1}{x-2} \int_{6}^{f(x)} 2tdt$.

First, let's evaluate the definite integral $\int_{6}^{f(x)} 2tdt$: $\int 2tdt = t^2$. So, $\int_{6}^{f(x)} 2tdt = [t^2]_{6}^{f(x)} = (f(x))^2 - 6^2 = (f(x))^2 - 36$.

Now, substitute this back into the limit expression: $L = \lim_{x\to3} \frac{(f(x))^2 - 36}{x-2}$.

We need to evaluate this limit. Let's examine the behavior of the numerator and the denominator as $x \to 3$. The denominator is $x-2$. As $x \to 3$, the denominator approaches $3-2 = 1$. The numerator is $(f(x))^2 - 36$. Since $f$ is a differentiable function, it is also continuous. Thus, $\lim_{x\to3} f(x) = f(3)$. We are given that $f(3) = 6$. So, as $x \to 3$, the numerator approaches $(f(3))^2 - 36 = 6^2 - 36 = 36 - 36 = 0$.

The limit is of the form $\frac{0}{1}$. $L = \frac{\lim_{x\to3} ((f(x))^2 - 36)}{\lim_{x\to3} (x-2)} = \frac{0}{1} = 0$.

The limit is 0.