Ivan Waters

2023-03-24

How to evaluate the limit $\frac{\mathrm{sin}\left(5x\right)}{x}$ as x approaches 0?

Tomas Cobb

Beginner2023-03-25Added 7 answers

Given $\underset{x\to 0}{lim}\frac{\mathrm{sin}\left(5x\right)}{x}$

We want to use the previously mentioned limit, so we need to have $\theta =5x$

To get 5x in the denominator, we'll multiply by $\frac{5}{5}$

$\underset{x\to 0}{lim}\frac{\mathrm{sin}\left(5x\right)}{x}=\underset{x\to 0}{lim}\frac{5\mathrm{sin}\left(5x\right)}{5x}$

Now, outside the limit, factor the 5 in the numerator.

$=5\underset{x\to 0}{lim}\frac{\mathrm{sin}\left(5x\right)}{5x}$

As $x\to 0$, $5x\to 0$ so we have:

$=5\left(1\right)=5$

(The limit is 5.)

Hence, using $\underset{\theta \to 0}{lim}\frac{\mathrm{sin}\theta}{\theta}=1$ and some other tools.

We want to use the previously mentioned limit, so we need to have $\theta =5x$

To get 5x in the denominator, we'll multiply by $\frac{5}{5}$

$\underset{x\to 0}{lim}\frac{\mathrm{sin}\left(5x\right)}{x}=\underset{x\to 0}{lim}\frac{5\mathrm{sin}\left(5x\right)}{5x}$

Now, outside the limit, factor the 5 in the numerator.

$=5\underset{x\to 0}{lim}\frac{\mathrm{sin}\left(5x\right)}{5x}$

As $x\to 0$, $5x\to 0$ so we have:

$=5\left(1\right)=5$

(The limit is 5.)

Hence, using $\underset{\theta \to 0}{lim}\frac{\mathrm{sin}\theta}{\theta}=1$ and some other tools.

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