(a) Let f(x)=sinx and an=n sin1 / n. Show that limn →∞ an-f^'(0)=1 (b) Let f(x) be differentiabl

step 1 Okay, here we are given a sequence with respect to our function f. Okay, suppose fx is equal to

(a) Let f(x)=sinx and an=n sin1 / n. Show that limn →∞...

(a) Let $f(x)=\sin x$ and $a_{n}=n \sin 1 / n$. Show that $\lim _{n \rightarrow \infty} a_{n}-f^{\prime}(0)=1$ (b) Let $f(x)$ be differentiable on the interval $[0,1]$ and $f(0)=0 .$ Consider the sequence $\left\{a_{n}\right\}$, where $a_{n}=n f(1 / n) .$ Show that $\lim _{n \rightarrow \infty} a_{n}=f^{\prime}(0)$.

(a) Let $f(x)=\sin x$ and $a_{n}=n \sin 1 / n$. Show that
$\lim _{n \rightarrow \infty} a_{n}-f^{\prime}(0)=1$
(b) Let $f(x)$ be differentiable on the interval $[0,1]$ and $f(0)=0 .$ Consider the sequence $\left\{a_{n}\right\}$, where $a_{n}=n f(1 / n) .$ Show that $\lim _{n \rightarrow \infty} a_{n}=f^{\prime}(0)$.

Key Concepts

Definition of the Derivative

The derivative at a point is defined as the limit of the difference quotient. In this context, since f is differentiable and f(0)=0, f'(0) can be obtained as the limit of f(x)/x as x approaches 0. This is a foundational concept that connects the behavior of a function near a point to the linear approximation at that point.

Limit of a Sequence

Understanding how a sequence behaves as its index approaches infinity is crucial here. By identifying the limit of the sequence a?, one can relate discrete approximations to continuous derivatives. This involves techniques that evaluate the convergence of terms defined as functions of 1/n.

Small Angle Approximation

For functions like sin?x, when x is close to zero, the small angle approximation (sin?x ? x) is used to simplify the analysis. This approximation underlies the proof of the limit lim(x?0)(sin?x)/x = 1, making it easier to relate the sequence terms to the derivative of sin?x at 0.

Transcript

00:01 Okay, here we are given a sequence with respect to our function f.

00:06 Okay, suppose fx is equal to sine x.

00:11 We want to define a n to be just equal to n times sine 1 over n.

00:18 For the first part of our question, we are required to show the limit when n approaches positive infinity of a n will just equal to the derivative of f at 0, which will be equal to 1.

00:40 Okay, notice our f is a trigonometric function.

00:46 It's just a simple sine x, so we know it's differentiable everywhere along the real line.

00:53 And at the point 0, we know f prime of 0 is defined as a limit when delta goes to 0.

01:05 Fx plus delta minus f of 0 divided by delta.

01:15 Okay, as f is equal to sine x, we know this limit can be expressed as limit when delta goes to 0.

01:24 F of, no, now x is equal to 0, so be equal to sine of delta minus sine 0, which is just equal to 0, divided by delta.

01:39 So it is actually equal to limit when delta goes to 0, sine delta divided by delta.

01:50 Okay, we know this limit always exists.

01:54 And this limit exists means for any delta.

02:04 Once delta goes to 0, we know this limit must exist.

02:08 So we can just let delta to be 1 over n, and the limit becomes the limit when n goes to positive infinity, sine 1 over n divided by 1 over n, and it is equivalent to saying this limit is just equal to f prime of 0.

02:36 Okay, for this limit, we know for any way, for any ways that delta approaches 0, this limit always exists.

02:52 Here we just let delta to be just equal to 1 over n...

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