Prepared by Professor Antonino De Martino (Polytechnic University of Milan) and Dr. Luana Manfredini.

Using Fundamental Limits

Theoretical Background

Here are the most important notable limits in calculus:

\[\lim_{x\to 0} \frac{\sin x}{x} = 1, \quad \lim_{x\to \infty} \frac{\sin x}{x} = 0, \quad \lim_{x\to 0} \frac{\log(1+x)}{x} = 1,\] \[\lim_{x\to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}, \quad \lim_{x\to \pm\infty} \left(1 + \frac{1}{x}\right)^x = e, \quad \lim_{x\to 0} (1+x)^{1/x} = e,\] \[\lim_{x\to 0} \frac{e^x - 1}{x} = 1, \quad \lim_{x\to 0} \frac{\tan x}{x} = 1, \quad \lim_{x\to 0} \frac{\arcsin x}{x} = 1, \quad \lim_{x\to 0} \frac{\arctan x}{x} = 1.\]

Key Theorems

Theorem: Non-Existence via Sequences

If there exist two sequences $ a_n \to c $, $ 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) \leq g(x) \leq h(x) $ near $ x = c $, and: \(\lim_{x \to c} f(x) = \lim_{x \to c} h(x) = L,\) then: \(\lim_{x \to c} g(x) = L.\)

Exercises

Example 1

\(\lim_{x \to +\infty} \left( \sqrt{x^2 + x + 1} - \sqrt{x^2 - x + 1} \right)\)

Example 2

\(\lim_{x \to \infty} x \log\left(\frac{x + 4}{x + 5}\right)\)

Example 3

\(\lim_{x \to \infty} \left(\frac{2x+9}{2x+1}\right)^x\)

Example 4

\(\lim_{x \to \infty} x \log\left(\frac{x^2 + 1}{x^2 + x}\right)\)

Example 5

\(\lim_{x \to \infty} \frac{\log(x^3 + 1)}{x}\)

Example 6

\(\lim_{x \to \infty} \frac{\sin x - x}{\cos x + \sqrt{1 + x^2}}\)

Example 7

\(\lim_{x \to \infty} \sin x \cdot \left[ \log(\sqrt{x} + 1) - \log(\sqrt{x + 1}) \right]\)

Example 8

\(\lim_{x \to \infty} \left(\frac{x + 3}{x - 1}\right)^{x + 1}\)

Example 9

\(\lim_{x \to 0^+} x^{1/\log(3x)}\)

Example 10

\(\lim_{x \to 0} \frac{e^x - e^{-x}}{x}\)

Solutions

Exercise 1

\[\lim_{x \to +\infty} \left( \sqrt{x^2 + x + 1} - \sqrt{x^2 - x + 1} \right)\]

Solution:

We multiply and divide by the conjugate expression:

\[\begin{aligned} \lim_{x \to +\infty} &\left( \sqrt{x^2 + x + 1} - \sqrt{x^2 - x + 1} \right) \\ &= \lim_{x \to +\infty} \frac{(x^2 + x + 1) - (x^2 - x + 1)}{\sqrt{x^2 + x + 1} + \sqrt{x^2 - x + 1}} \\ &= \lim_{x \to +\infty} \frac{2x}{\sqrt{x^2 + x + 1} + \sqrt{x^2 - x + 1}} \\ &= \lim_{x \to +\infty} \frac{2x}{x \left( \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + \sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} \right)} \\ &= \frac{2}{1 + 1} = 1 \end{aligned}\]

Exercise 2

\[\lim_{x \to \infty} x \cdot \log\left( \frac{x + 4}{x + 5} \right)\]

Solution:

We rewrite the logarithmic expression:

\[\lim_{x \to \infty} x \cdot \log\left( 1 - \frac{1}{x + 5} \right)\]

Let \(t = -\frac{1}{x + 5} \), so\) x = \frac{-1 - 5t}{t} $$. Then:

\[\lim_{t \to 0} \frac{-1 - 5t}{t} \cdot \log(1 + t) = -1\]

Exercise 3

\[\lim_{x \to +\infty} \left( \frac{2x + 9}{2x + 1} \right)^x\]

Solution:

We write:

\[\left( \frac{2x + 9}{2x + 1} \right)^x = \left( 1 + \frac{8}{2x + 1} \right)^x\]

Let \(t = \frac{8}{2x + 1}\), so \(x = \frac{4}{t} - \frac{1}{2}\). Then:

\[(1 + t)^{\frac{4}{t} - \frac{1}{2}} = \frac{(1 + t)^{4/t}}{(1 + t)^{1/2}} \to \frac{e^4}{1} = e^4\]

Exercise 4

\[\lim_{x \to +\infty} x \cdot \log\left( \frac{x^2 + 1}{x^2 + x} \right)\]

Solution:

We simplify the ratio:

\[\frac{x^2 + 1}{x^2 + x} = \frac{1 + \frac{1}{x^2}}{1 + \frac{1}{x}}\]

Let \(t = \frac{1}{x}\), so \(x = \frac{1}{t}\). Then:

\[\lim_{t \to 0} \frac{1}{t} \cdot \log\left( \frac{1 + t^2}{1 + t} \right) = \lim_{t \to 0} \left( \frac{\log(1 + t^2)}{t} - \frac{\log(1 + t)}{t} \right) = -1\]

Exercise 5

\[\lim_{x \to +\infty} \frac{\log(x^3 + 1)}{x}\]

Solution:

We use the identity:

\[\log(x^3 + 1) = \log\left(x^3\left(1 + \frac{1}{x^3}\right)\right) = 3 \log x + \log\left(1 + \frac{1}{x^3} \right)\]

So:

\[\frac{3 \log x + \log\left(1 + \frac{1}{x^3} \right)}{x} = \frac{3 \log x}{x} + \frac{\log\left(1 + \frac{1}{x^3} \right)}{x} \to 0 + 0 = 0\]

Exercise 6

\[\lim_{x \to +\infty} \frac{\sin x - x}{\cos x + \sqrt{1 + x^2}}\]

Solution:

We rewrite the expression:

\[\frac{\sin x - x}{\cos x + \sqrt{1 + x^2}} = \frac{\frac{\sin x}{x} - 1}{\frac{\cos x}{x} + \sqrt{1 + \frac{1}{x^2}}}\]

As \(x \to +\infty\), we know:

\[\frac{\sin x}{x} \to 0, \quad \frac{\cos x}{x} \to 0\]

So the limit becomes:

\[\frac{0 - 1}{0 + 1} = -1\]

Exercise 7

\[\lim_{x \to +\infty} \sin x \cdot \left[ \log(\sqrt{x} + 1) - \log(\sqrt{x + 1}) \right]\]

Solution:

We use the identity:

\[\log(\sqrt{x} + 1) - \log(\sqrt{x + 1}) = \log\left( \frac{\sqrt{x} + 1}{\sqrt{x + 1}} \right)\]

Now:

\[\frac{\sqrt{x} + 1}{\sqrt{x + 1}} = \frac{\sqrt{x}(1 + \frac{1}{\sqrt{x}})}{\sqrt{x} \cdot \sqrt{1 + \frac{1}{x}}} = \frac{1 + \frac{1}{\sqrt{x}}}{\sqrt{1 + \frac{1}{x}}}\]

As \(x \to +\infty\), this tends to 1, so the logarithm tends to 0. Since \(\sin x\) is bounded, the product tends to 0:

\[\lim_{x \to +\infty} \sin x \cdot \log\left( \frac{\sqrt{x} + 1}{\sqrt{x + 1}} \right) = 0\]

Exercise 8

\[\lim_{x \to +\infty} \left( \frac{x + 3}{x - 1} \right)^{x + 1}\]

Solution:

We rewrite:

\[\left( \frac{x + 3}{x - 1} \right)^{x + 1} = \left( 1 + \frac{4}{x - 1} \right)^{x + 1} = \left( 1 + \frac{4}{x - 1} \right)^x \cdot \left( 1 + \frac{4}{x - 1} \right)\]

Let \(y = \frac{4}{x - 1}\), so \(x = \frac{4 + y}{y}\). Then:

\[\lim_{y \to 0} (1 + y)^{\frac{4}{y}} \cdot (1 + y) = e^4\]

Exercise 9

\[\lim_{x \to 0^+} x^{\frac{1}{\log(3x)}}\]

Solution:

Let \(y = \frac{1}{\log(3x)}\), then \(x = \frac{e^{1/y}}{3}\), so:

\[\lim_{y \to 0^+} \left( \frac{e^y}{3} \right)^y = \frac{e^{y \cdot \frac{1}{y}}}{3^y} = \frac{e}{1} = e\]

Exercise 10

\[\lim_{x \to 0} \frac{e^x - e^{-x}}{x}\]

Solution:

We split the expression:

\[\frac{e^x - e^{-x}}{x} = \frac{e^x - 1}{x} + \frac{1 - e^{-x}}{x} = \frac{e^x - 1}{x} + \frac{e^{-x} - 1}{-x}\]

Using the known limits:

\[\lim_{x \to 0} \frac{e^x - 1}{x} = 1, \quad \lim_{x \to 0} \frac{e^{-x} - 1}{-x} = 1\]

So the total limit is:

\(1 + 1 = 2\)