Limit of cos(x)/x and sec(x)/x

[Jhenrique]

Homework Statement

Compute: $$\lim_{x \to 0} = \frac{\cos(x)}{x}$$ $$\lim_{x \to 0} = \frac{\cosh(x)}{x}$$ $$\lim_{x \to 0} = \frac{\sec(x)}{x}$$ $$\lim_{x \to 0} = \frac{sech(x)}{x}$$

Homework Equations

$$\lim_{x \to x_0} = \frac{f(x)}{g(x)} = \lim_{x \to x_0} = \frac{\frac{df}{dx}(x)}{\frac{dg}{dx}(x)}$$

The Attempt at a Solution

$$\lim_{x \to 0} = \frac{\cos(x)}{x} = \lim_{x \to 0} \frac{ \frac{d}{dx}\cos(x)}{\frac{d}{dx}x} = \lim_{x \to 0} \frac{- \sin(x)}{1} = 0$$ $$\\\lim_{x \to 0} = \frac{\cosh(x)}{x}=0 \\ \\\lim_{x \to 0} = \frac{\sec(x)}{x}=0 \\ \\\lim_{x \to 0} = \frac{sech(x)}{x}=0$$ Correct?

 

[jbunniii]

The Attempt at a Solution

$$\lim_{x \to 0} = \frac{\cos(x)}{x} = \lim_{x \to 0} \frac{ \frac{d}{dx}\cos(x)}{\frac{d}{dx}x} = \lim_{x \to 0} \frac{- \sin(x)}{1} = 0$$

You can't apply L'Hopital's rule unless you have an indeterminate form. What is $\lim_{x \rightarrow 0} \cos(x)$?

 

[Jhenrique]

You can't apply L'Hopital's rule unless you have an indeterminate form. What is $\lim_{x \rightarrow 0} \cos(x)$?

Is 1

 

[micromass]

You can easily check what the solution is by simply graphing the function. That will give you a big hint. Of course a graph is not a proof, but it helps.

 

[Mark44]

Minor point, but you're showing all your limits incorrectly.

This --
$$\lim_{x \to 0} = \frac{\cos(x)}{x}$$
-- should be written without the = between "lim" and the function you're taking the limit of.

 

[Jhenrique]

Minor point, but you're showing all your limits incorrectly.

This --
$$\lim_{x \to 0} = \frac{\cos(x)}{x}$$
-- should be written without the = between "lim" and the function you're taking the limit of.

I didn't even realize! LOOOOOOL

 

[Curious3141]

L' Hopital's Rule can only be used for limits of the form 0/0 or infinity over infinity (though in the latter case, the signs of the infinity don't matter). There are other criteria, but if your limit doesn't even fit into one of those indeterminate forms, LHR doesn't apply.