Question

If \(\lim _{\mathrm{x} \rightarrow 0}[\\{\operatorname{Sin}[(\mathrm{n}+1) \mathrm{x}]+\sin \mathrm{x}\\} / \mathrm{x}]=(1 / 2)\) then value of n is: (a) \(-2.5\) (b) \(-0.5\) (c) \(-1.5\) (d) \(-1\)

Short answer

The value of n is \(-0.5\), which corresponds to option (b).

Step-by-step solution

Rewrite the given expression using the sine addition formula

Since we have a sine of the sum inside the limit, we can use the sine addition formula to rewrite the expression. The sine addition formula is: \(\sin(A + B) = \sin A \cos B + \cos A \sin B\) Applying this formula to \(\sin[(n+1)x]\), where \(A = nx\) and \(B = x\), we get: \(\sin[(n+1)x] = \sin nx \cos x + \cos nx \sin x\) Now, we can rewrite the expression inside the limit as: \(\frac{\sin nx \cos x + \cos nx \sin x + \sin x}{x}\)

Split the expression into two fractions

We can split the expression into two fractions by separating the terms with sine x: \(\frac{\sin nx \cos x + \cos nx \sin x + \sin x}{x} = \frac{\sin nx \cos x}{x} + \frac{\cos nx \sin x + \sin x}{x}\)

Factor out sin x from the second fraction

We can factor out \(\sin x\) from the second fraction: \(\frac{\sin nx \cos x}{x} + \frac{\cos nx \sin x + \sin x}{x} = \frac{\sin nx \cos x}{x} + \frac{\sin x (\cos nx + 1)}{x}\)

Apply limit properties and trigonometric limits

Now, we can apply the limit properties and some known trigonometric limits: \(\lim_{x \rightarrow 0} \frac{\sin nx \cos x + \sin x (\cos nx + 1)}{x} = \lim_{x \rightarrow 0} \frac{\sin nx \cos x}{x} + \lim_{x \rightarrow 0} \frac{\sin x (\cos nx + 1)}{x}\) We know that \(\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1\), so we can use this limit to evaluate our expression: \(\lim_{x \rightarrow 0} \frac{\sin nx \cos x}{x} = n \lim_{x \rightarrow 0} \frac{\sin x}{x} \cos x = n\) \(\lim_{x \rightarrow 0} \frac{\sin x (\cos nx + 1)}{x} = (\lim_{x \rightarrow 0} \frac{\sin x}{x})(\lim_{x \rightarrow 0} (\cos nx + 1)) = 1\) Now, we can rewrite the given limit as: \(n + 1 = \frac{1}{2}\)

Solve for n

Finally, we can solve for n: \(n = \frac{1}{2} - 1\) \(n = -\frac{1}{2}\) So, the value of n is \(-0.5\), which corresponds to option (b).

Key concepts

Sine Addition Formula

The sine addition formula is a fundamental identity in trigonometry, which states that for any two angles, A and B, the sine of their sum can be expressed as:

\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]

Understanding this formula is crucial in solving trigonometric expressions involving the sum of angles.

For example, when encountering a limit problem involving sine functions with angle sums, as seen in the given exercise, we can apply the sine addition formula to break down the expression into simpler terms that can be handled more easily. This step is essential because it transforms the original trigonometric function into forms that are often finalized by utilizing basic trigonometric limits.

Trigonometric Limits

A vital aspect in calculus is the comprehension of trigonometric limits, which are the building blocks for analyzing the behavior of trigonometric functions as the angle tends to a specific value.

One of the classic trigonometric limits that students learn is:

\[ \lim_{{x \to 0}} \frac{{\sin x}}{x} = 1 \]

Knowing this fundamental limit simplifies the process of evaluating limits involving sine functions, as evident in the provided exercise. When we encounter expressions like \( \frac{{\sin nx}}{x} \) within a limit heading to zero, we can directly apply this principle to ascertain the result without complex computations. It's essential to recognize these patterns to operate limits efficiently. Failure to apply these limits can result in unnecessarily complicated calculations, possibly leading to errors or misconceptions regarding the limit's value.

L'Hopital's Rule

When dealing with more complex limits that result in indeterminate forms such as \( 0/0 \) or \( \infty/\infty \), L'Hopital's rule becomes a powerful tool.

It states that if:\[ \lim_{{x \to c}} f(x) = 0 \] and \[ \lim_{{x \to c}} g(x) = 0 \] or both limits approach infinity, then if the derivative of f and g exist and \( g'(x) \) is not zero near c, we can find the limit of \( f(x)/g(x) \) as x approaches c by:

\[ \lim_{{x \to c}} \frac{{f(x)}}{{g(x)}} = \lim_{{x \to c}} \frac{{f'(x)}}{{g'(x)}} \]

If the right-hand side limit exists or is infinity, then it is the value of the original limit. This rule is indispensable, notably when standard trigonometric limit principles do not straightforwardly apply. However, in the given example of a trigonometric limit involving \( \sin x/x \), L'Hopital's rule was not necessary, because the limit could be determined directly using basic trigonometric limits.