I wouldn't think you need to use "L'Hopital". The only reason you need to look close is because both numerator and denominator are 0 when x= 4. And, since these are polynomials, that means that x-4 is a factor of both.
\(\displaystyle x^2- 6x+ 8= (x- 4)(x- 2)\) and
\(\displaystyle x^2- 7x+ 12= (x- 4)(x- 3)\) so
\(\displaystyle \lim_{x\to 4}\frac{x^2- 6x+ 8}{x^2- 7x+ 12}= \lim_{x\to 4}\frac{x- 2}{x- 3}= \frac{4- 2}{4- 3}= 2\).
Then, in order to have continuity at x= 4 we must have \(\displaystyle \lim_{x\to 4} 2e^{x^2- 3x+ c}= 2\)
\(\displaystyle \lim_{x\to 4} 2e^{x^2- 3x+ c}= 2e^{16- 6+ c}= 2e^{10+ c}= 2\).
So we must have \(\displaystyle e^{10+ c}= 1\) which means we must have 10+ c= 1 or c= -9.