$$\lim_{x \to 49} {\sqrt{x}-7 \over x-49}$$
Using my properties of limits I just plugged in $49$ for $x$ and ended up with an answer of $0$. I was wondering if I did this correctly as it almost feels too simple to be correct
$\endgroup$ 63 Answers
$\begingroup$For $x\neq 49$ and $x>0$, $$\dfrac{\sqrt{x}-7}{x-49}=\dfrac{\sqrt{x}-7}{\sqrt{x}^2-7^2}=\dfrac{\sqrt{x}-7}{(\sqrt{x}-7)(\sqrt{x}+7)}=\dfrac{1}{\sqrt{x}+7}$$
So the function can be prolongated in $x=49$ and the limit is $1/(\sqrt{49}+7)=1/14$.
Remarks on indeterminate "0/0" $f:x\mapsto \frac{x^2}{x}$ can be made continuous in 0 (with $f(0)=0$). $f:x\mapsto \frac{x}{x^2}$ cannot be made continuous in 0. $f:x\mapsto \frac{x}{x}$ can be made continuous in 0 (with $f(0)=1$). The question is whether the function has a limit around the point where it is not defined.
$\endgroup$ $\begingroup$An indeterminate form "$0/0$" does not mean the limit does not exist. It can exist (think of for instance $\frac{\sin x}{x}$ when $x\to 0$), can fail to exist (e.g., $\frac{\lvert x\rvert}{x^2}$ when $x\to 0$) -- everything is possible, so you need to dig deeper to find the answer.
Hint: can you recognize a derivative i.e. something of the form $$ f'(a) = \lim_{x\to a} \frac{f(x)-f(a)}{x-a} $$ for some function $f$ differentiable at $a$ (here, $a=49$)?
$\endgroup$ 7 $\begingroup$Rationalizing the numerator is the usual method for this, and someone's already posted that. Here's another method. Let $u = \sqrt x$, so $x = u^2$ and as $x\to49$ then $u\to 7$. Then you have: $$ \lim_{x\to49} \frac{\sqrt x - 7}{x-49} = \lim_{u\to7} \frac{u - 7}{u^2 - 49} = \lim_{u\to7} \frac{u-7}{(u-7)(u+7)} = \cdots. $$
IMPORTANT: The most important reason for learning about limits is to deal with limits of quotients in which the numerator and denominator both approach $0$. That's because the derivative $\dfrac{dy}{dx}$ is defined as $$ \frac{dy}{dx} = \lim_{\Delta x\to0} \frac{\Delta y}{\Delta x} $$ and the numerator and denominator both approach $0$.
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