4.0 Some Standard Limits

  Limits

• $\mathop {\lim }\limits_{x \to 0} \tfrac{{\sin x}}{x} = 1$, Where $x$ in radians
• $\mathop {\lim }\limits_{x \to 0} \tfrac{{\tan x}}{x} = 1$, Where $x$ in radians
• $\mathop {\lim }\limits_{x \to 0} {(1 + x)^{1/x}} = e$
• $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^x} = e$
• $\mathop {\lim }\limits_{x \to 0} \tfrac{{{a^x} - 1}}{x} = \ln (a),a \in {R^ + }$
• $\mathop {\lim }\limits_{x \to 0} \tfrac{{{e^x} - 1}}{x} = 1$
• $\mathop {\lim }\limits_{x \to 0} \tfrac{{{{(1 + x)}^n} - 1}}{x} = n$
• $\mathop {\lim }\limits_{x \to 0} \tfrac{{\ln (1 + x)}}{x} = 1$
• $\mathop {\lim }\limits_{x \to a} \left( {\frac{{{x^n} - {a^n}}}{{x - a}}} \right) = n.{a^{n - 1}}$, where $n$ is a rational number.
• $\mathop {\lim }\limits_{x \to 0} \tfrac{{{{\sin }^{ - 1}}x}}{x} = 1$
• $\mathop {\lim }\limits_{x \to 0} \tfrac{{{{\tan }^{ - 1}}x}}{x} = 1$
• $\mathop {\lim }\limits_{x \to 0} \tfrac{{\log _a (1 + x)}}{x} = {\log _a}e,a > 0, \ne 1$

Question 9.

Find $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{x}$.

Solution:

As $x \to 0$, $ax \to 0$.

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{x}$ $ = \mathop {\lim }\limits_{ax \to 0} \frac{{\sin ax}}{{ax}} \times a$ [Dividing numerator and denominator with a]

$ = a\mathop {\lim }\limits_{ax \to 0} \frac{{\sin ax}}{{ax}}$

$ = a\mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{t}$ [putting $ax = t$. As $ax \to 0$, $t \to 0$ ]

$ = a$ [using Standard Limit $\mathop {\lim }\limits_{x \to 0} \tfrac{{\sin x}}{x} = 1$]

Question 10.

Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{{x^2}}}$.

Solution:

$\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{{x^2}}} \times \frac{{1 + \cos x}}{{1 + \cos x}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\cos }^2}x}}{{{x^2}(1 + \cos x)}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x}}{{{x^2}(1 + \cos x)}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x}}{{{x^2}}} \times \mathop {\lim }\limits_{x \to 0} \frac{1}{{1 + \cos x}}$

$ = 1 \times \frac{1}{{1 + 1}} = \frac{1}{2}$