Theoretical Recall
The main notable limits in calculus are:
\[\lim_{x\to 0}\frac{\sin x}{x}=1, \quad \lim_{x\to \infty}\frac{\sin x}{x}=0,\] \[\lim_{x\to 0}\frac{\log(1+x)}{x}=1, \quad \lim_{x\to 0}\frac{1-\cos x}{x^{2}}=\tfrac{1}{2},\] \[\lim_{x\to +\infty}\left(1+\tfrac{1}{x}\right)^{x}=e, \quad \lim_{x\to -\infty}\left(1+\tfrac{1}{x}\right)^{x}=e,\] \[\lim_{x\to 0}(1+x)^{1/x}=e, \quad \lim_{x\to 0}\frac{e^{x}-1}{x}=1,\] \[\lim_{x\to 0}\frac{\tan x}{x}=1, \quad \lim_{x\to 0}\frac{\arcsin x}{x}=1, \quad \lim_{x\to 0}\frac{\arctan x}{x}=1.\]Key Theorems
Theorem (Non-existence via sequences).
If there exist two sequences $a_{n},b_{n}\to c$ such that
\(\lim_{n\to\infty}f(a_{n}) \ne \lim_{n\to\infty}f(b_{n}),\)
then $\lim_{x\to c}f(x)$ does not exist.
Theorem (Squeeze Theorem).
If $f(x)\le g(x)\le h(x)$ and
\(\lim_{x\to c}f(x)=\lim_{x\to c}h(x)=L,\)
then
\(\lim_{x\to c}g(x)=L.\)
Exercise 1
\(\lim_{x\to +\infty}\Big(\sqrt{x^{2}+x+1}-\sqrt{x^{2}-x+1}\Big).\)
Solution:
Rationalize the numerator:
\(\frac{(\sqrt{x^{2}+x+1}-\sqrt{x^{2}-x+1})(\sqrt{x^{2}+x+1}+\sqrt{x^{2}-x+1})}{\sqrt{x^{2}+x+1}+\sqrt{x^{2}-x+1}}.\)
Simplify: \(\frac{2x}{x\left(\sqrt{1+\tfrac{1}{x}+\tfrac{1}{x^{2}}}+\sqrt{1-\tfrac{1}{x}+\tfrac{1}{x^{2}}}\right)} \to 1.\)
Final Result:
\(1\)
Exercise 2
\(\lim_{x\to \infty}x\log\!\Big(\tfrac{x+4}{x+5}\Big).\)
Solution:
Rewrite:
\(x\log\!\Big(1-\tfrac{1}{x+5}\Big).\)
Let $t=-\tfrac{1}{x+5}$, so $x=\tfrac{-1-5t}{t}$. Then: \(\lim_{t\to 0}\frac{-1-5t}{t}\log(1+t)=-1.\)
Final Result:
\(-1\)
Exercise 3
\(\lim_{x\to +\infty}\left(\tfrac{2x+9}{2x+1}\right)^{x}.\)
Solution:
Rewrite:
\(\Big(1+\tfrac{8}{2x+1}\Big)^{x}.\)
Let $t=\tfrac{8}{2x+1}$, then $x=\tfrac{4}{t}-\tfrac{1}{2}$. So: \(\lim_{t\to 0}(1+t)^{4/t-1/2}=\frac{(1+t)^{4/t}}{(1+t)^{1/2}}\to e^{4}.\)
Final Result:
\(e^{4}\)
Exercise 4
\(\lim_{x\to +\infty}x\log\Big(\tfrac{x^{2}+1}{x^{2}+x}\Big).\)
Solution:
Let $t=1/x$, so $x=1/t$. Then:
\(\lim_{t\to 0}\frac{1}{t}\log\Big(\tfrac{1+t^{2}}{1+t}\Big).\)
This equals: \(\lim_{t\to 0}\frac{\log(1+t^{2})}{t}-\frac{\log(1+t)}{t}=-1.\)
Final Result:
\(-1\)
Exercise 5
\(\lim_{x\to +\infty}\frac{\log(x^{3}+1)}{x}.\)
Solution:
Rewrite:
\(\frac{3\log x+\log(1+\tfrac{1}{x^{3}})}{x}.\)
As $x\to\infty$, numerator grows like $\log x$, denominator like $x$. Limit is 0.
Final Result:
\(0\)
Exercise 6
\(\lim_{x\to +\infty}\frac{\sin x -x}{\cos x+\sqrt{1+x^{2}}}.\)
Solution:
Divide numerator and denominator by $x$:
\(\frac{\tfrac{\sin x}{x}-1}{\tfrac{\cos x}{x}+\sqrt{1+\tfrac{1}{x^{2}}}}.\)
As $x\to\infty$, $\tfrac{\sin x}{x}\to 0$, $\tfrac{\cos x}{x}\to 0$. So limit is $-1$.
Final Result:
\(-1\)
Exercise 7
\(\lim_{x\to +\infty}\left(\frac{x+3}{x-1}\right)^{x+1}.\)
Solution:
Rewrite:
\(\left(1+\tfrac{4}{x-1}\right)^{x+1}.\)
Let $y=\tfrac{4}{x-1}$, so $x=\tfrac{4+y}{y}$. Then: \(\lim_{y\to 0}(1+y)^{4/y}(1+y)=e^{4}.\)
Final Result:
\(e^{4}\)
Exercise 8
\(\lim_{x\to 0^{+}}x^{1/\log(3x)}.\)
Solution:
Let $y=1/\log(3x)$, so $x=e^{1/y}/3$. Then:
\(\lim_{y\to 0^{+}}\left(\tfrac{e^{y}}{3}\right)^{y}=\lim_{y\to 0^{+}}\tfrac{e}{3^{y}}=e.\)
Final Result:
\(e\)
Exercise 9
\(\lim_{x\to 0}\frac{e^{x}-e^{-x}}{x}.\)
Solution:
Rewrite:
\(\frac{e^{x}-1}{x}+\frac{e^{-x}-1}{-x}.\)
Each tends to 1 as $x\to 0$. Sum is 2.
Final Result:
\(2\)
Exercise 10
\(\lim_{x\to 1}\frac{\cos\!\left(\tfrac{\pi}{2}x\right)}{x-1}.\)
Solution:
Let $y=x-1$, $x=y+1$. Then:
\(\lim_{y\to 0}\frac{\cos(\tfrac{\pi}{2}y+\tfrac{\pi}{2})}{y}
=\lim_{y\to 0}-\frac{\sin(\tfrac{\pi}{2}y)}{y}.\)
So: \(-\frac{\pi}{2}.\)
Final Result:
\(-\tfrac{\pi}{2}\)