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Kryger [21]
3 years ago
8

Evaluate the limit with either L'Hôpital's rule or previously learned methods.lim Sin(x)- Tan(x)/ x^3x → 0

Mathematics
1 answer:
Vsevolod [243]3 years ago
3 0

Answer:

\dfrac{-1}{6}

Step-by-step explanation:

Given the limit of a function expressed as \lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}, to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

= \dfrac{sin(0)-tan(0)}{0^3}\\= \frac{0}{0} (indeterminate)

Step 2: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the function

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\

Step 3: substitute x = 0 into the resulting function

= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)

Step 4: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\

=  \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)

Step 6: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\

Step 7: substitute x = 0 into the resulting function in step 6

=  \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}

<em>Hence the limit of the function </em>\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \  is \ \dfrac{-1}{6}.

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After applying <em>algebraic</em> analysis we find the <em>right</em> choices for each case, all of which cannot be presented herein due to <em>length</em> restrictions. Please read explanation below.

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In this problem we have a rational function, whose features can be inferred by algebraic handling:

Holes - x-values that do not belong to the domain of the <em>rational</em> function:

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But one root is an evitable discontinuity as:

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Vertical asymptotes - There is a <em>vertical</em> asymptote where a hole exists. Hence, the function has two vertical asymptotes.

Horizontal asymptotes - <em>Horizontal</em> asymptote exists and represents the <em>end</em> behavior of the function if and only if the grade of the numerator is not greater than the grade of the denominator. If possible, this assymptote is found by this limit:

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Lastly, we have the following conclusions:

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  3. This function has x-intercepts? True
  4. One <em>vertical</em> asymptote along the line where x always equals what number: 1
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To learn more on rational functions: brainly.com/question/27914791

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