Compute $\lim _{x\to \:k}\frac{s\cdot x^2y\cdot sin\left(k-x\right)}{k^2-kx}$ without using L'Hopitals Rule

Question

How do I compute $\lim _{x\to \:k}\frac{s\cdot x^2y\cdot sin\left(k-x\right)}{k^2-kx}$ without using L'Hopitals Rule.

$\lim _{x\to \:k}\frac{s\cdot x^2y\cdot sin\left(k-x\right)}{k^2-kx}$

I know $\lim _{x\to \:0}\frac{sin\left(x\right)}{x}=1$, but how do I change $\lim _{x\to \:k}\frac{s\cdot x^2y\cdot sin\left(k-x\right)}{k^2-kx}$ to $\lim _{x\to \:0}\frac{sin\left(x\right)}{x}=1$

Answer 1

Hint:

\begin{align} \lim _{x\to k}\frac{s\cdot x^2y\cdot \sin\left(k-x\right)}{k^2-kx}&=\left(\lim_{x\to k}\frac{sx^2y}{k}\right)\left(\lim_{x\to k}\frac{\sin \left(k-x\right)}{k-x}\right) \end{align} Since the last two limits exist, in particular by setting $u=k-x$ we get $$\lim_{x\to k}\frac{\sin \left(k-x\right)}{k-x}=\lim_{u\to 0}\frac{\sin u}{u}=1$$

Answer 2

You haven't told us what $s,y$ are. I assume they are constants, which brings up the question: Why include them?

Hint: The only thing that matters here is

$$\frac{\sin (k-x)}{k^2 - kx} = \frac{\sin (k-x)}{k(k - x)} = \frac{1}{k}\frac{\sin (k-x)}{(k - x)}.$$

As $x\to k, k-x \to 0,$ therefore _____ .

Answer 3

Let $x-k =z$. Then $(s x^2 y \sin (k-x))/(k^2-kx)=( s x^2 y k^{-1}) ((\sin z)/z) $.And $z\to 0$ as $x\to k$.

Answer 4

Factor $k$ in the denominator of $\lim _{x\to \:k}\frac{s\cdot x^2y\cdot sin\left(k-x\right)}{k^2-kx}$.

Do you see what to do from here?

$\lim \:_{x\to k}\frac{s\cdot \:x^2y\cdot \:sin\left(k-x\right)}{k^2-kx}=\lim \:\:_{x\to k}\frac{s\cdot \:\:x^2y\cdot \:\:sin\left(k-x\right)}{k\left(k-x\right)}=\lim \:\:\:_{x\to \:k}\frac{s\cdot \:\:\:x^2y\:\:}{k}\cdot \lim \:\:\:\:_{x\to \:\:k}\frac{\:sin\left(k-x\right)}{\left(k-x\right)}$

I will assume you can do the rest.