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Chapter 4: Problem 27

Evaluate the following limits. $$\lim _{x \rightarrow 0} \frac{e^{x}-\sin x-1}{x^{4}+8 x^{3}+12 x^{2}}$$

Short Answer

Expert verified
Function: $$\lim _{x \rightarrow 0} \frac{e^{x}-\sin x-1}{x^{4}+8 x^{3}+12x^{2}}$$ Answer: The limit of the given function as x approaches 0 is $$\frac{1}{24}$$ using L'H么pital's Rule.

Step by step solution

01

Identify the indeterminate form

First, let's plug in x = 0 into the given function to see if it is in an indeterminate form: $$\frac{e^0-\sin0-1}{0^4+8(0)^3+12(0)^2}=\frac{1-0-1}{0}=\frac{0}{0}$$ Since the function is in the form 0/0, we can use L'H么pital's Rule.
02

Find the first derivatives of the numerator and denominator

In order to apply L'H么pital's Rule, we need to find the first derivative of both the numerator and the denominator with respect to x: Numerator: $$\frac{d}{dx}(e^x-\sin x-1)=e^x-\cos x$$ Denominator: $$\frac{d}{dx}(x^4+8x^3+12x^2)=4x^3+24x^2+24x$$
03

Evaluate the limit with the first derivatives

Now, let's evaluate the new limit after finding the first derivatives using L'H么pital's Rule: $$\lim _{x \rightarrow 0} \frac{e^{x}-\cos x}{4x^{3}+24x^{2}+24x}$$ Plug in x = 0: $$\frac{e^0-\cos0}{4(0)^3+24(0)^2+24(0)}=\frac{1-1}{0}=\frac{0}{0}$$ Again, we have an indeterminate form 0/0, so we must apply L'H么pital's Rule once more.
04

Find the second derivatives of the numerator and denominator

We need to find the second derivative of both the numerator and the denominator with respect to x: Numerator: $$\frac{d^2}{dx^2}(e^x-\cos x)=e^x+\sin x$$ Denominator: $$\frac{d^2}{dx^2}(4x^3+24x^2+24x)=12x^2+48x+24$$
05

Evaluate the limit with the second derivatives

Now, let's evaluate the new limit after finding the second derivatives using L'H么pital's Rule: $$\lim _{x \rightarrow 0} \frac{e^{x}+\sin x}{12x^{2}+48x+24}$$ Plug in x = 0: $$\frac{e^0+\sin0}{12(0)^2+48(0)+24}=\frac{1+0}{24}=\frac{1}{24}$$ Now we have a limit that is not in an indeterminate form.
06

Conclusion

The final answer is: $$\lim _{x \rightarrow 0} \frac{e^{x}-\sin x-1}{x^{4}+8 x^{3}+12x^{2}}=\frac{1}{24}$$

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