Solving limits that approach infinity

Solve limit: $$\lim_{k\to\infty}\frac{4k+3}{2}$$ My approach. $$\lim_{k\to\infty}\frac{(4k+3)}{2} \frac{1/k}{1/k}$$ $$\lim_{k\to\infty}\frac{4+3/k}{2/k}$$ Then, we know that the limit of k as k approaches inf for $$\frac{3}{k}$$ goes to 0 and the same goes for $$\frac{2}{k}$$ however for both limits, they can also be viewed as infinitesmall such that the final answer is $$\frac{4}{0+} = \infty$$

Can somebody verify if I did this correctly? I'm confused as to whether the fraction limits approach 0 or infinitesmall because if it's 0 then my limit is undefined

• You have limits for $n$, but the arguments are in terms of $k$. – Benedict W. J. Irwin Jul 26 '18 at 15:26
• Sry I fixed it to be variable k. – Donald Devy Jul 26 '18 at 15:28
• Isn't it obvious from the beginning that $4k+3\to\infty$ as $k\to\infty?$ – saulspatz Jul 26 '18 at 15:29
• True... I actually didn't have to do anything. But for curiosity's sake can you help me with my understanding for the fraction limits? – Donald Devy Jul 26 '18 at 15:31
• You did this correctly (as in you are using heuristics which can be made rigorous). You should dispense of your worries with the limit begin undefined; the limit must be infinite, since $(4k+3)/2$ becomes arbitrarily large as $k$ does. – Mike Earnest Jul 26 '18 at 15:55

Simply note that

$$\frac{4k+3}{2}=2k+\frac32\ge 2k \ge k \to \infty$$

then conclude by squeeze theorem.

Why so complicated? $$\lim_{k\to\infty}\frac{4k+3}{2}=\lim_{k\to\infty}(2k+1.5)=2\infty+1.5=\infty$$

• While right in spirit, treating $\infty$ as a number gives me the willies. – Randall Jul 26 '18 at 15:34
• As long as you don't do things like $\infty/\infty$ or $\infty\cdot 0$ or $\infty-\infty$, you should be fine. It will boil down to the answer form @gimusi – Andrei Jul 26 '18 at 15:36

Can't you find $k$ such that $$\frac{4k+3}2>1000000\ ?$$

And larger numbers ?

If yes, the expression is unbounded.