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Hello find the limit stated below

[tex]\\ \rm\Rrightarrow {\displaystyle{\lim_{x\to 0}}}\dfrac{sin^3x+3cosx-cot^2x}{tan^3x+7sinx+3sec^2x}-e^{3x}[/tex]


Note:-

Make sure to include proper explanation and all steps ​

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Answer:

[tex]\displaystyle \lim_{x \to 0} \frac{\sin^3 x + 3\cos x - \cot^2 x}{\tan^3 x + 7 \sin x + 3 \sec^2 x} - e^{3x} = \boxed{- \infty}[/tex]

General Formulas and Concepts:
Calculus

Limits

  • Graphical Limits

Right-Side Limit:
[tex]\displaystyle \lim_{x \to c^+} f(x)[/tex]

Left-Side Limit:
[tex]\displaystyle \lim_{x \to c^-} f(x)[/tex]

Limit Rule [Variable Direct Substitution]:
[tex]\displaystyle \lim_{x \to c} x = c[/tex]

Discontinuities

  • Removable (Hole)
  • Jump
  • Infinite (Asymptote)

Step-by-step explanation:

Step 1: Define

Identify given.

[tex]\displaystyle \lim_{x \to 0} \frac{\sin^3 x + 3\cos x - \cot^2 x}{\tan^3 x + 7 \sin x + 3 \sec^2 x} - e^{3x}[/tex]

Step 2: Find Limit

Let's start out by directly substituting x = 0 into our limit.

  1. [Limit] Apply Limit Rule [Variable Direct Substitution]:
    [tex]\displaystyle \lim_{x \to 0} \frac{\sin^3 x + 3\cos x - \cot^2 x}{\tan^3 x + 7 \sin x + 3 \sec^2 x} - e^{3x} = \frac{\sin^3 0 + 3\cos 0 - \cot^2 0}{\tan^3 0 + 7 \sin 0 + 3 \sec^2 0} - e^{3(0)}[/tex]
  2. Simplify:
    [tex]\displaystyle \lim_{x \to 0} \frac{\sin^3 x + 3\cos x - \cot^2 x}{\tan^3 x + 7 \sin x + 3 \sec^2 x} - e^{3x} = \text{unde} \text{fined}[/tex]

We can see that the cotangent of 0 is undefined. Since we do not have an indeterminate form, we cannot use L'Hopital's Rule. Let's look at the function graphically (refer to the attachment).

Approaching the left limit, we can see that the graph heads towards negative infinity.

Approaching the right limit, we can see that the graph also heads towards negative infinity.

∴ we can say that the limit as x approaches 0 is equal to negative infinity.

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Learn more about limits: https://brainly.com/question/23339726

Learn more about calculus: https://brainly.com/question/27351658

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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

Ver imagen Space
LammettHash

It's plain to see that [tex]e^{3x}\to1[/tex] as [tex]x\to0[/tex], so I'll focus on the trigonometric expression.

Multiply uniformly through the expression by [tex]\sin^2(x)\cos^3(x)[/tex] :

[tex]\dfrac{\sin^3(x) + 3\cos(x) - \cot^2(x)}{\tan^3(x) + 7\sin(x) + 3\sec^2(x)}[/tex]

[tex]= \dfrac{\cos^3(x) \left(\sin^5(x) + 3\cos(x)\sin^2(x) - \cos^2(x)\right)}{\sin^2(x) \left(\sin^3(x) + 7\cos^3(x)\sin(x) + 3\cos(x)\right)}[/tex]

Now as [tex]x\to0[/tex],

• [tex]\cos^3(x)\to1[/tex]

• [tex]\sin^5(x)+3\cos(x)\sin^2(x)-\cos^2(x)\to-1[/tex]

• [tex]\sin^3(x) + 7\cos^3(x)\sin(x) + 3\cos(x) \to 3[/tex]

but [tex]\sin^2(x)\to0[/tex], so [tex]x = 0[/tex] a non-removable discontinuity. Then from either side of [tex]x = 0[/tex], [tex]\frac1{\sin^2(x)}\to+\infty[/tex], and taking into account the sign from the second point above, the overall limit would be

[tex]\displaystyle \lim_{x\to0} \frac{\sin^3(x) + 3\cos(x) - \cot^2(x)}{\tan^3(x) + 7\sin(x) + 3\sec^2(x)} - e^{3x} = -\infty-1 = \boxed{-\infty}[/tex]

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