On one of my past calculus tests, I struggled with a multi-part question that asked to "Evaluate the limits that exist, or prove that they don't", as I would often be plagued by doubt wondering if each limit exists, and wound up wasting time changing my answers.
Now with my final upcoming later this week, the last thing I want to do is spend any time second-guessing myself as to whether or not the limit exists. What sort of techniques can I use to intuitively determine the existence of a limit of a function before delving into evaluation or a proof of non-existence?
I'm in a first year single variable Calculus course; here are a few limits.
$$\lim_{x \to 0} \frac{\tan(3x^2) + \sin^2(5x)}{x^2}$$
$$\lim_{x \to 3^+} \frac{\sqrt{x – 3}}{|x – 3|}$$
$$\lim_{x \to 1} \frac{x^2 – \sqrt{x}}{x – 1}$$
Best Answer
You want to be on the lookout for familiar limits, relating new limits to limits you have experience with.
The first limit can be manipulated so that it is in terms of famous limits: notice that $$\frac{\tan(3x^2)}{x^2} = 3\left(\frac{\tan(3x^2)}{3x^2}\right) = 3\left(\frac{\sin(3x^2)}{3x^2}\right)\left(\frac{1}{\cos(3x^2)}\right)$$ and $$\frac{\sin^2(5x)}{x^2} = \left(\frac{\sin(5x)}{x}\right)\left(\frac{\sin(5x)}{x}\right) = 25\left(\frac{\sin(5x)}{5x}\right)\left(\frac{\sin(5x)}{5x}\right).$$ When $x\to 0$, the cosine gives no trouble, and the limits involving the sine are of the form $$\lim_{u\to 0}\frac{\sin(u)}{u}$$ which you should recognize. This suggests that the limit exists, and how to compute it.
The second limit is obfuscated: since $x\to 3^+$, you can drop the absolute value sign (since $x\gt 3$ so $|x-3| = x-3$) and then you have $$\frac{\sqrt{x-3}}{|x-3|} = \frac{\sqrt{x-3}}{x-3}$$ which is of the form $\frac{\sqrt{u}}{u}$. It should now be easy to see that as $u\to 0^+$, this limit does not exist (simplify it).
For the third limit, you have a polynomial in $\sqrt{x}$ divided by a polynomial in $\sqrt{x}$, and both go to $0$ when $x\to 1$. You should expect to be able to factor out $(\sqrt{x}-1)$ from both numerator and denominator and cancel them. Since the denominator is $(\sqrt{x}-1)(\sqrt{x}+1)$, there will be only a single factor of $\sqrt{x}-1$, so the limit should exist after you cancel those factors (the new denominator will not be $0$ at $x=1$). Indeed: the numerator is $u^4 - u = u(u^3-1)=u(u-1)(u^2+u+1)$ (with $u=\sqrt{x}$), the denominator is $(u-1)(u+1)$, so you can cancel and evaluate.