Cfrgtkky

Consider the function-
$$g(x)=begin{cases} 1, text{ if } xin[-1,1]\ 0, text{ otherwise } end{cases}$$

and $$f(x)=lim_{hto0}frac{1}{2h}int_{x-h}^{x+h}g(y)dy$$

then what is the value of $f(1)?$

My attempt:

We get, $$f(1)=lim_{hto0}frac{1}{2h}int_{1-h}^{1+h}g(y)dy$$

Applying Newton-Leibnitz for the nuemerator after applying L'Hopital for this $0/0$ limit, we get

$$f(1)=frac{g(1+h)+g(1-h)}{2}=1$$

Am I correct?

Since the function under limit is an even function of $h$ it is sufficient to deal with $hto 0^{+}$ only.

We have $$lim_{hto 0^{+}}frac{1}{2h}int_{1-h}^{1+h}g(y),dy=lim_{hto 0^{+}}frac{1}{2h}int_{1-h}^{1}g(y),dy$$ The integral evaluates to $h$ and hence the above limit is $1/2$.

Your approach is fine but the issue is that you seem to assume $g$ continous at $1$.

You are replacing the wrong variable in your solution.

$$
f(1)=lim_{hto 0}frac{1}{2h}int_{1-h}^{1+h}g(y)dy.
$$

Now, for each fixed $h<2$,

$$
int_{1-h}^{1+h}g(y)dy=1-(1-h)=h.
$$

Thus,

$$
f(1)=lim_{hto 0}frac{h}{2h}=lim_{hto 0}frac{1}{2}=frac{1}{2}.
$$

$$f(1)=lim_{hto0}frac{1}{2h}int_{1-h}^{1+h}g(y)dy = lim_{hto0}frac{1}{2h}int_{1-h}^{1}g(y)dy = lim_{hto0}frac{1}{2h}(1-(1-h)) ={1over 2}$$

The two sided limit, $lim_{h rightarrow 0}~g(1+h)$, does not exist because $g(x)$ is discontinuous in $x=1$. (To be more specific, we are dealing with a jump discontinuity.) This will cause problems when computing $f(1)$.

If we would to take the one-sided limit from above, then we would get $lim_{h rightarrow 0^+}~g(1+h) = 0$ and $lim_{h rightarrow 0^+}~g(1-h) = 1$.

I have not done the full derivation but I am guessing the final answer would be $f(1)=1/2$.

I hope this helps you out enough, good luck solving the math problem!