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$$\lim_{x \to 0}(x\sec x)$$ So putting in $x=0$ you get the answer $0$. $$\lim_{x \to 0}(x\sec x)=0$$ My question is is this a correct way to solve?

edit : So from the answers below, I've understood that if a function is continuous, then $\lim_{x \to a}f(x)=f(a)$

But how do you figure out if a function is continuous? from the graph? But what if it's a function that I don't know the graph of?

Raknos13
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Yes it is the correct approach , since the function is continuous , specially at $x=0$.

Remember , quotients, products, sums of continuous functions remain continuous. (As long as there is no domain problem)

You can just safely put $x=0$ directly without having to check for left hand and right hand limits separately. Also , since plugging $x=0$ does not cause any domain problems , you're good to go !

$$\lim_{x \to 0}(x\sec x)=0$$

This can also be seen as -

$\lim_{x \to 0}(x\cdot{sec(x)})=\lim_{x \to 0}\left(\frac{x}{cos(x)}\right)$

Now using the McLaurin series for Cosine function , we have

$$\lim_{x \to 0}\left(\frac{x}{1-\frac{x^2}{2!}+\frac{x^4}{4!}\cdots}\right)=0$$

Tanuj
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Yes. For continuous functions, you can just plug in the value, because you can switch a limit and a continuous function.

I.e., if $g$ is continuous, then $$\lim_{x \to a} g(f(x)) = g \left(\lim_{x\to a}f(x)\right)$$

provided the limit on the right exists and gives a value on which $g$ is defined.

  • How do you find out whether a certain given function is continuous or discontinuous? For example, if it's a common function that I know the graph of I can, but what if I don't know the graph of a certain function? – Raknos13 Dec 26 '17 at 14:11
  • @mettledmike This can be taken as a rule in general , a continuous function at a point multiplied by another continuous function at that same point results in a continuous function at that point. Also , I have edited my answer , see if it helps you . – Tanuj Dec 26 '17 at 14:22
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    It should be standard knowledge that quotients, products, sums of continuous functions remain continuous. For the sec function, you know it is the reciprocal of a well known continuous function (i.e. cosine function), so it is also continuous. –  Dec 26 '17 at 14:27
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$$\lim_{x \to 0}(x\sec x)=\lim_{x \to 0}x\cdot\frac{1}{\cos x}=\lim_{x \to 0}\frac{x}{\cos x}=0$$

Adi Dani
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It is correct as long as you can justify it. You have theorems such that:

  • $f (x)=x $ is continuous on the whole $\mathbb R $
  • $f (x)=\cos x $ is continuous on the whole $\mathbb R$
  • If two functions $f (x ) $ and $g (x) $ are continuous at $x=x_0$ and $g (x_0)\ne 0$, then the function $h (x)=\frac {f (x)}{g (x)} $ is also continuous at $x=x_0$.

Using all those, we prove that $f (x)=x\sec x=\frac {x}{\cos x} $ is continuous at $x=0$, which by definition of continuity means that $\lim_{x\to 0}f (x)=f (0) $, and so you can substitute $x=0$ instead of calculating the limit.

To re-iterate, it is the continuity of the function $x\sec x $ at $x=0$ (which we can prove) that allows you to do that substitution. Beware other problems, where it may not hold!