## MA101.11 Finding Limits – Examples

A number of examples of how to find limits. \begin{aligned} \displaystyle \lim_{x\to 1} \{ x^2 + \frac{x^3+1}{x^2 +1} + 2x\sqrt{1+x^2} \} &= 1 + \frac{1+1}{1+1}+2\cdot 1\cdot\sqrt{1+1} \\&= 2 + 2\sqrt{2} \end{aligned} \begin{aligned} \displaystyle \lim_{x\to \infty} \frac{x^2-3x+2}{2x^2+4x-7} &= \lim_{x\to \infty} \frac{1 - \frac{3}{x} + \frac{2}{x^2}}{2 + \frac{4}{x} - \frac{7}{x^2}} \\&= \frac{1}{2} \end{aligned} \begin{aligned} \displaystyle \lim_{x\to 2} \frac{x^2 - 4}{x^2-5x-6} &= \lim_{x\to 2} \frac{(x-2)(x+2)}{(x-2)(x-3)} \\&= \lim_{x\to 2} \frac{(x+2)}{(x-3)} \\&= -4 \end{aligned} \begin{aligned} \displaystyle \lim_{x\to \infty} (\sqrt{x^2+6x+4}-x) &= \lim_{x\to \infty} \frac{(\sqrt{x^2+6x+4}-x)(\sqrt{x^2+6x+4}+x)}{(\sqrt{x^2+6x+4}+x)} \\&= \lim_{x\to \infty} \frac{x^2 + 6x + 4 - x^2}{(\sqrt{x^2+6x+4}+x)} \\&= \lim_{x\to \infty} \frac{6+\frac{4}{x}}{\sqrt{1+\frac{6}{x}+\frac{4}{x^2}}+1} \\&= \frac{6}{\sqrt{1}+1} = 3 \end{aligned} \begin{aligned} \displaystyle \lim_{x\to 1^+} \frac{\sqrt{x-1}}{x^2-1} &= \lim_{x\to 1^+} \frac{1}{\sqrt{x-1}(x+1)} \\&= \infty \end{aligned} \begin{aligned} \displaystyle \lim_{x\to 0} \frac{x}{\sqrt{1-x}-\sqrt{1+x}} &= \lim_{x\to 0} \frac{x(\sqrt{1-x}+\sqrt{1+x})}{(1-x)-(1+x)} \\&= \lim_{x\to 0} \frac{x(\sqrt{1-x}+\sqrt{1+x}}{-2x} \\&= -1 \end{aligned} \begin{aligned} \displaystyle \lim_{x\to 1} \frac{ln x}{x^2-1} &= \lim_{x\to 1} \frac{\frac{1}{x}}{2x} \:\mbox{ - (Using L'Hopital's Rule)} \\&= \frac{1}{2} \end{aligned} \begin{aligned} \displaystyle \lim_{x\to 0} \frac{sin(2x)}{x} &= \lim_{x\to 0} \frac{2cos(2x)}{1} \\&= 1 \end{aligned} \begin{aligned} \displaystyle \lim_{x\to 0} \frac{1-cos(x)}{x^2} &= \lim_{x\to 0} \frac{sin(x)}{2x} \\&= \frac{1}{2} \end{aligned} \begin{aligned} \displaystyle \lim_{x\to 0} (cos(x))^\frac{1}{x} &= \lim_{x\to 0} e^{ln(cos(x)^\frac{1}{x}} \\&= \lim_{x\to 0} e^{\frac{1}{x}ln(cos(x))} \\&= e^{\displaystyle\lim_{x\to 0} \frac{ln(cos(x))}{x}} \end{aligned}

Now, \begin{aligned} \displaystyle \lim_{x\to 0} \frac{ln(cos(x))}{x} &= \lim_{x\to 0} \frac{\frac{-sin(x)}{cos(x)}}{1} \\&= \lim_{x\to 0} -tan(x) = 0 \end{aligned}

So, $\displaystyle \lim_{x\to 0} (cos(x))^\frac{1}{x} = 1$ \begin{aligned} \displaystyle \lim_{x\to 0^+} x lnx &= \lim_{x\to 0^+} \frac{lnx}{\frac{1}{x}} \\&= \lim_{x\to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} \\&= \lim_{x\to 0^+} -x = 0 \end{aligned} \begin{aligned} \displaystyle \lim_{x\to \infty} x^3 e^{-x^2} &= \lim_{x\to \infty} \frac{x^3}{e^{x^2}} \\&= \lim_{x\to \infty} \frac{3x^2}{2x e^{x^2}} \\&= \lim_{x\to \infty} \frac{3}{4x e^{x^2}} = 0 \end{aligned} \begin{aligned} \displaystyle \lim_{x\to \infty} (1+\frac{1}{x})^x &= \lim_{x\to \infty} e^{ln(1+\frac{1}{x})^x} \\&= \lim_{x\to \infty} e^{xln(1+\frac{1}{x})} \\&= e^\lim_{x\to \infty} xln(1+\frac{1}{x})} \end{aligned

Now, \begin{aligned} \displaystyle \lim_{x\to \infty} x ln(1 + \frac{1}{x}) &= \lim_{x\to \infty} \frac{ln(1+\frac{1}{x})}{\frac{1}{x}} \\&= \lim_{x\to \infty} \frac{(-\frac{1}{x^2}\cdot\frac{1}{1+\frac{1}{x}})}{-\frac{1}{x^2}} \\&= \lim_{x\to \infty} \frac{1}{1+\frac{1}{x}} = 1 \end{aligned}

So, $\displaystyle \lim_{x\to \infty} (1+\frac{1}{x})^x = e^1 = e$

## MA101.10 How to find limits

If a function is know to be  continuous at $x=a$, then by the definition of continuity we have, $\displaystyle\lim_{t\to a} f(x) = f(a)$.

We will take is as known that the elementary functions – polynomial, rational (polynomial/polynomial), trigonometrical, exponential, logarithmic) – are continuous everywhere.

Therefore, although not very exciting, we are able to say such things as: $\displaystyle\lim_{x\to \frac{\pi}{2}} sin x = sin \frac{\pi}{2}=1$ $\displaystyle\lim_{x\to 3} e^x = e^3$ $\displaystyle\lim_{x\to 2} \frac{x(x-1)}{x-3} = \frac{2(2-1)}{2-3} = -2$

A very useful rule (it’s a theorem really) for finding limits is L’Hôpitals rule (LH), which is really a collection of rules. The most comprehensive form of the rule is as follows:

Suppose you wish to find something like $\displaystyle\lim_{x\to[\;]} \frac{f(x)}{g(x)}$ where [ ] may be $a, a^+, a^-, \infty, -\infty$.

Suppose further that $\displaystyle\lim_{x\to[\;]} f(x) = \lim_{x\to [\;]} g(x) = \{\}$ where either $\{\}=0$  – ( $\frac{0}{0}$ type), or $\{\}=\infty$ – ( $\frac{\infty}{\infty}$ type).

Then the LH rule says look at $\displaystyle\lim_{x\to [\;]} \frac{f'(x)}{g'(x)}$ and if you can do it, say find it to be ( ) then the original limit is the same.

Danger: Do not use LH when it is not applicable.

For example $\displaystyle\lim_{x\to 0} \frac{sin(x) + 1}{sin(x)+2} = \frac{1}{2}$ is not the same as $\displaystyle\lim_{x\to 0} \frac{cos(x)}{cos(x)} = 1$.

The LH rule gives us with no effort some useful results. For example: $\displaystyle\lim_{x\to 0} \frac{sin(x)}{x} = \lim_{x\to 0} \frac{cos(x)}{1} = 1$