To compute the limit
[tex]\displaystyle\lim_{x\to0}\frac{e^x-\sin x-1}{x^4+7x^3+8x^2}[/tex]
notice that when x = 0, both the numerator and denominator converge to 0. Apply l'Hopital's rule once gives
[tex]\displaystyle\lim_{x\to0}\frac{e^x-\cos x}{4x^3+21x^2+16x}[/tex]
but again, this returns the indeterminate form 0/0. Applying again gives
[tex]\displaystyle\lim_{x\to0}\frac{e^x+\sin x}{12x^2+42x+16}[/tex]
so that the numerator converges to e⁰ + sin(0) = 1, and the denominator converges to 12•0² + 42•0 + 16 = 16.
So the limit is 1/16.
