Filtering property of Dirac $\delta$ function 
Evaluate $$\int_0^\infty f(x)\delta(x-1)dx$$where$$f(x)=\begin{cases}x^2,&0\le x<1\\\sin 2,&x=1\\x,&x>1\end{cases}$$

Attempt
Since the function is discontinuous at $1$, I couldn't directly say the answer would be $f(1)$. I considered the $\delta$-sequence$$\delta_k(x-1)=\begin{cases}\frac k2,& |x-1|<1/k\\0,&\text{otherwise}\end{cases}\\\int_0^\infty f(x)\delta_k(x-1)dx=\frac k2\Big[\int_{1-1/k}^1x^2dx+\int^{1+1/k}_1xdx\Big]\\=1+\frac12\Big(\frac1{3k^2}-\frac1{2k}\Big)$$
Then taking the limit as $k\to\infty$, I got the answer as $1$, which seems correct intuitively as it is the limit of the function at $x=1$. But the answer key gives the answer $\sin 2=f(1)$. 
Is the key wrong or am I missing something?
Edit
The book defines the Dirac $\delta$ function (also called impulse function) $\delta(t)$ as the limit of a sequence of functions $\{\delta_k(t)\}$, where $$\int_{-\infty}^\infty\delta_k(t)dt=1$$For example,$$\delta_k(t)=\begin{cases}\frac k2,&|t|<1/k\\0,&\text{otherwise}\end{cases}$$is one such sequence. Then it goes on to say that the Dirac $\delta$ function can be understood as the generalized function$$\delta(t)=\begin{cases}0,&t\ne0\\\infty,&t=0\end{cases}$$For continuous functions in $[0,\infty)$ and $a\ge0$,$$\int_0^\infty f(t)\delta(t-a)=f(a)$$
 A: The function $f$ has a removable discontinuity at $1$.  Therefore, we have
$$\langle \delta_{1},f\rangle=\lim_{x\to 1}f(x)$$
To see this, we denote $\delta_n(x)$ as a regularization of the Dirac Delta, which for any suitable test function, $\phi(x)$, satisfies 
$$\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x)\phi(x)\,dx=\phi(0)$$
Now, suppose we have a function $f(x)$ that is of compact support and is smooth everywhere except at $1$ where it has a removable discontinuity.  Let $g(x)$ be defined as 
$$g(x)=\begin{cases}f(x)&,x\ne 1\\\\\lim_{x\to 1}f(x)&, x=1\end{cases}$$
Then, since $g$ is a suitable test functions we see that 
$$\begin{align}
\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x-1)f(x)\,dx&=\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x-1)g(x)\,dx\\\\
&=g(1)\\\\
&=\lim_{x\to 1}f(x)
\end{align}$$
And we are done!

EXAMPLE:
As an example, suppose $\delta_n(x)$ is the asymmetrically centered pulse function, which for $a\in(0,1)$ given by
$$\delta_n(x)=\begin{cases}n&,x\in[-a/n,(1-a)/n]\\\\0&,\text{otherwise}\end{cases}$$
Then, we have 
$$\begin{align}
\lim_{n\to\infty}\int_{-\infty}^\infty \delta_n(x-1)f(x)\,dx&=\lim_{n\to\infty}\left(n\int_{1-a/n}^{1+(1-a)/n} f(x)\,dx\right)\\\\
&=\lim_{n\to\infty}\left(n\int_{1-a/n}^{1} f(x)\,dx\right)+\lim_{n\to\infty}\left(n\int_{1}^{1+(1-a)/n} f(x)\,dx\right)\\\\
&=a\lim_{x\to 1^-}f(x)+(1-a)\lim_{x\to 1^+}f(x)\\\\
&=\lim_{x\to 1}f(x)
\end{align}$$
since the discontinuity at $1$ is removable and therefore the right-side and left-side limits are equal.

NOTE:
It is of paramount importance to understand that if $f$ has a removable discontinuity at $x_0$, then the functional $\langle \delta_{x_0},f\rangle =\lim_{x\to x_0}f(x)$ but if $f$ has a jump discontinuity at $x_0$, then the functional $\langle \delta_{x_0},f\rangle$ is not defined.

In fact, I showed in This Answer, that if $H$ is he Heaviside Function, then $\langle \delta_0,H\rangle $ is meaningless.

