Evaluate $\lim_{k\to l}\frac{\sqrt{e^{\sin(2k)}}-\sqrt{e^{\cos(2k)}}}{k-l}$ without L'Hopital

Question

Evaluate $$\lim_{k\to l}\frac{\sqrt{e^{\sin(2k)}}-\sqrt{e^{\sin(2l)}}}{k-l}$$

I know this is equivalent with $\frac{d}{dx}\sqrt{e^{\sin(2x)}}=\cos(2x)\sqrt{e^{\sin(2x)}}$, but my instructor wants me to solve this using limit. My attempt:

Multiply with $\frac{\sqrt{e^{\sin(2k)}}+\sqrt{e^{\sin(2l)}}}{\sqrt{e^{\sin(2k)}}+\sqrt{e^{\sin(2l)}}}$

$$ \frac{1}{2}\lim_{k\to l}\frac{e^{\sin(2k)}-e^{\sin(2l)}}{(k-l)(\sqrt{e^{\sin(2l)}})}$$

Here's where I stuck.

Answer 1

\begin{align*} \lim_{k\to l}\frac{\sqrt{e^{\sin(2k)}}-\sqrt{e^{\sin(2l)}}}{k-l} &=\lim_{k\to l}\frac{e^{\sin(2k)}-e^{\sin(2l)}}{(\sqrt{e^{\sin(2k)}}+\sqrt{e^{\sin(2l)}})(k-l)}\\ &=\frac1{2\sqrt{e^{\sin(2l)}}}\lim_{k\to l}\frac{e^{\sin(2k)}-e^{\sin(2l)}}{k-l}\\ &=\frac{\sqrt{e^{\sin(2l)}}}{2}\lim_{k\to l}\frac{e^{\sin(2k)-\sin(2l)}-1}{k-l}\\ &=\frac{\sqrt{e^{\sin(2l)}}}{2}\lim_{k\to l}\frac{e^{2\cos(k+l)\sin(k-l)}-1}{2\cos(k+l)\sin(k-l)}\cdot\frac{\sin(k-l)}{k-l}\cdot 2\cos(k+l)\\ &=\cos(2l)\sqrt{e^{\sin(2l)}}\left(\lim_{x\to0}\frac{e^x-1}{x}\right)\left(\lim_{x\to0}\frac{\sin x}x\right)\\ &=\cos(2l)\sqrt{e^{\sin(2l)}}. \end{align*} The without-L'Hospital proof for last two limits can be found here and here.

Answer 2

Hint: Your last line must be $$\frac{e^{\sin(2k)}-e^{\sin(2l)}}{(k-l)(\sqrt{e^{\sin(2k}}+\sqrt{e^{\sin(2l)}})}$$

Answer 3

$$\lim_{k\to l}\frac{\sqrt{e^{\sin(2k)}}-\sqrt{e^{\sin(2l)}}}{k-l} =\lim_{k\to l}\frac1{\sqrt{e^{\sin(2k)}}+\sqrt{e^{\sin(2l)}}}\cdot\lim_{k\to l}\frac{e^{\sin(2k)}-e^{\sin(2l)}}{k-1}$$

Now $$\lim_{k\to l}\frac{e^{\sin(2k)}-e^{\sin(2l)}}{k-1}=e^{\sin(2l)}\cdot\lim_{k\to l}\dfrac{e^{(\sin2k-\sin2l)}-1}{\sin2k-\sin2l}\cdot\lim_{k\to l}\dfrac{\sin2k-\sin2l}{k-l}$$

Use Prosthaphaeresis Formulas, $\sin2k-\sin2l=2\sin(k-l)\cos(k+l)$