Algebra of Limits

Algebra of Limits

Let f and g be two real functions with domain D. We define four new functions f ± g , f/g, fg on domain D by setting (f ± g) (x) = f(x) ± g(x), (fg) (x) = f(x) g(x)

(f/g) (x) = f(x)/g(x), if g(x) ≠ 0 for any x ϵ D

1. Let \(\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=l\) and \(\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)=m\). If l and m exist, then

• \(\underset{x\to a}{\mathop{\lim }}\,\left( f\pm g \right)\left( x \right)=\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\pm \,\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)=l\pm m\),
• \(\underset{x\to a}{\mathop{\lim }}\,\left( fg \right)\left( x \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)=lm\),
• \(\underset{x\to a}{\mathop{\lim }}\,\left( \frac{f}{g} \right)\left( x \right)=\frac{\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)}{\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)}=\frac{l}{m}\) Provided.

2. \(\underset{x\to a}{\mathop{\lim }}\,Kf\left( x \right)=K.\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\) Where K is constant.

3. \(\underset{x\to a}{\mathop{\lim }}\,\left| f\left( x \right) \right|=\left| \underset{x\to a}{\mathop{\lim }}\,f\left( x \right) \right|=\left| l \right|\).

4. \(\underset{x\to a}{\mathop{\lim }}\,{{\left[ f\left( x \right) \right]}^{g\left( x \right)}}={{l}^{m}}\).

5. f(x) ≤ g(x) for every x in the deleted nbd of a, then \(\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\le \underset{x\to a}{\mathop{\lim }}\,g\left( x \right)\).

6. f(x) ≤ g(x)  ≤ hfor every x in the deleted nbd of a and \(\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=l=\underset{x\to a}{\mathop{\lim }}\,h\left( x \right)\) then \(\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)=l\).

7. \(\underset{x\to a}{\mathop{\lim }}\,fog\left( x \right)=f\left( \underset{x\to a}{\mathop{\lim }}\,f\left( x \right) \right)=\log l\).

In particular:

1. \(\underset{x\to a}{\mathop{\lim }}\,\log f\left( x \right)=\log \left( \underset{x\to a}{\mathop{\lim }}\,f\left( x \right) \right)=\log l\).

2. \(\underset{x\to a}{\mathop{\lim }}\,{{e}^{f\left( x \right)}}={{e}^{\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)}}={{e}^{l}}\).

3. If \(\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\) = + or – ∞ then \(\underset{x\to a}{\mathop{\lim }}\,\frac{1}{f\left( x \right)}=0\).

Evaluation of limits: In the previous sections, we have discussed the notion of left hand limit (LHL), right hand limit (RHL) and the existence of the limit of a function f(x) at a given point. In the evaluation of limits, it is assumed that \(\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\) always exists i.e., \(\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)\).

In this section, we will discuss various methods of evaluating limits. To facilitate the job of evaluation of limits we categorize problems on limits in the following categories:

1. Algebraic limits
2. Trigonometric limits
3. Exponential and logarithmic limits
4. Various methods of evaluations of algebraic limits:

To evaluate algebraic limits, we have the following methods:

1. Direct substitution method
2. Factorization method
3. Rationalisation method
4. Using some standard results
5. Method of evaluating limits when variable tends to ∞ or -∞
6. Direct substitution method:

If by direct substitution of the point in the given expression we get a finite number, then the number obtained is the limit of the given expression.

Example: evaluate \(\underset{x\to 1}{\mathop{\lim }}\,\) (3x² + 4x + 5)

Solution:  \(\underset{x\to 1}{\mathop{\lim }}\,\) (3x² + 4x + 5) = 3(1)² + 4 (1) + 5 = 12

\(\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos x}{1+\sin x}\),

\(\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos x}{1+\sin x}=\frac{\cos 0}{1+\sin 0}=1\).

Factorisation Method: Consider \(\underset{x\to a}{\mathop{\lim }}\,\frac{f\left( x \right)}{g\left( x \right)}\). If by putting x = a the rational function \(\frac{f\left( x \right)}{g\left( x \right)}\) takes the form 0/0, ∞/∞ etc, then (x – a) is a factor of both f(x) and g(x) . In such a case we factories the numerator and denominator and then cancel out the common factor (x – a). After cancelling out the common factor x – a we again put x = a in the given expression and see whether we get a meaningful number or not. This process is repeated till we get a meaningful number.

Evaluate: \(\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{3}}-6{{x}^{2}}+11x-6}{{{x}^{2}}-6x+8}\).

Solution: we have,

\(\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{3}}-6{{x}^{2}}+11x-6}{{{x}^{2}}-6x+8}\),

\(=\underset{x\to 2}{\mathop{\lim }}\,\frac{\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)}{\left( x-2 \right)\left( x-4 \right)}\),

\(=\underset{x\to 2}{\mathop{\lim }}\,\frac{\left( x-1 \right)\left( x-3 \right)}{\left( x-4 \right)}\),

= ½

Rationalisation Method: This method is generally used when one of numerator and denominator or both of them consist of expressions involving square roots.

Evaluate: \(\underset{x\to a}{\mathop{\lim }}\,\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}\).

Solution: We have,

\(\underset{x\to a}{\mathop{\lim }}\,\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}\),

\(=\underset{x\to a}{\mathop{\lim }}\,\left( \frac{\left( \sqrt{a+2x}-\sqrt{3x} \right)\left( \sqrt{a+2x}+\sqrt{3x} \right)\left( \sqrt{3a+x}+2\sqrt{x} \right)}{\left( \sqrt{3a+x}-2\sqrt{x} \right)\left( \sqrt{3a+x}+2\sqrt{x} \right)\left( \sqrt{a+2x}+\sqrt{3x} \right)} \right)\),

\(=\underset{x\to a}{\mathop{\lim }}\,\frac{\left( a-x \right)\sqrt{3a+x}+2\sqrt{x}}{3\left( a-x \right)\left( \sqrt{a+2x}+\sqrt{3x} \right)}\),

\(=\underset{x\to a}{\mathop{\lim }}\,\frac{\left( \sqrt{3a+x}+2\sqrt{x} \right)}{3\left( \sqrt{a+2x}+\sqrt{3}x \right)}\),

\(=\frac{4\sqrt{a}}{2\left( 3\sqrt{3a} \right)}=\frac{2}{3\sqrt{3}}\).

Evaluate of Limits using the Result: \(\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}}\), where n ϵ Q.

Evaluate: \(\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{m}}-{{a}^{m}}}{{{x}^{n}}-{{a}^{n}}}\).

Solution: We have,

\(\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{m}}-{{a}^{m}}}{{{x}^{n}}-{{a}^{n}}}\),

\(=\underset{x\to a}{\mathop{\lim }}\,\left[ \frac{{{x}^{m}}-{{a}^{m}}}{x-a}.\frac{x-a}{{{x}^{n}}-{{a}^{n}}} \right]\),

\(=\underset{x\to a}{\mathop{\lim }}\,\left[ \frac{{{x}^{m}}-{{a}^{m}}}{x-a}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a} \right]\),

\(=\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{m}}-{{a}^{m}}}{x-a}\div \underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a}\),

\(=m{{a}^{m-1}}\div n{{a}^{n-1}}=\frac{m}{n}{{a}^{m-n}}\).