Math  /  Calculus

Question(1 point)
Evaluate the limit using L'Hospital's rule if necessary. limx+x16ex\lim _{x \rightarrow+\infty} \frac{x^{16}}{e^{x}}
Answer: \square

Studdy Solution

STEP 1

1. We are given the limit limx+x16ex\lim_{x \rightarrow +\infty} \frac{x^{16}}{e^{x}}.
2. We need to evaluate this limit as xx approaches positive infinity.
3. L'Hospital's rule can be applied if the limit is in an indeterminate form like \frac{\infty}{\infty}.

STEP 2

1. Determine if the limit is in an indeterminate form.
2. Apply L'Hospital's Rule if necessary.
3. Evaluate the limit after applying L'Hospital's Rule.
4. Conclude the value of the limit.

STEP 3

Check if the limit is in an indeterminate form:
As x+x \rightarrow +\infty, both the numerator x16x^{16} and the denominator exe^x approach \infty.
Thus, the limit is in the form \frac{\infty}{\infty}, which is an indeterminate form.

STEP 4

Apply L'Hospital's Rule:
L'Hospital's Rule states that if limxcf(x)g(x)\lim_{x \rightarrow c} \frac{f(x)}{g(x)} is in an indeterminate form, then:
limxcf(x)g(x)=limxcf(x)g(x)\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}
Differentiate the numerator and the denominator:
Numerator: ddx[x16]=16x15\frac{d}{dx}[x^{16}] = 16x^{15}
Denominator: ddx[ex]=ex\frac{d}{dx}[e^x] = e^x

STEP 5

Evaluate the new limit:
limx+16x15ex\lim_{x \rightarrow +\infty} \frac{16x^{15}}{e^x}
This is still in the form \frac{\infty}{\infty}, so apply L'Hospital's Rule again.
Differentiate again:
Numerator: ddx[16x15]=240x14\frac{d}{dx}[16x^{15}] = 240x^{14}
Denominator: ddx[ex]=ex\frac{d}{dx}[e^x] = e^x
Evaluate the new limit:
limx+240x14ex\lim_{x \rightarrow +\infty} \frac{240x^{14}}{e^x}
Continue applying L'Hospital's Rule repeatedly until the power of xx in the numerator is zero.

STEP 6

Continue applying L'Hospital's Rule 16 times in total:
After 16 applications, the numerator becomes a constant (the derivative of x0x^0) and the denominator remains exe^x.
The final limit is:
limx+16!ex=0\lim_{x \rightarrow +\infty} \frac{16!}{e^x} = 0
since exe^x grows much faster than any constant.
The value of the limit is:
0\boxed{0}

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