Answered

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Use I'Hopital's Rule more than once to rewrite the limit in its final form as lim x --> 0
lim x --> 0 e^x -sinx - 1/ x^4 + 7x^3 + 8x^2

Sagot :

LammettHash

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.

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