ONE MORE STEP TOWARD BETTER TOMORROW : chapter 14 Worked Out Examples 5

Saturday, 9 August 2014

chapter 14 Worked Out Examples 5

Example: 7

Evaluate the following limits:

	(a) $\mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{{a^x} + {b^x} + {c^x}}}{3}} \right)^{\dfrac{1}{x}}}$
	(b) $\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {\dfrac{\pi }{4} + x} \right)} \right)^{\dfrac{1}{x}}}$
	(c) $\mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{\sin x}}{x}} \right)^{\dfrac{1}{{{x^2}}}}}$
	(d) $\mathop {\lim }\limits_{x \to 0} {(\cos x)^{\dfrac{1}{{\sin x}}}}$
	(e) $\mathop {\lim }\limits_{x \to 0} {\left( {\cos x + a\sin bx} \right)^{\dfrac{1}{x}}}$

Solution: 7-(a)

Notice that all the limits above are of the form ${(f(x))^{g(x)}}$ where $f\left( x \right) \to 1$ and $g\left( x \right) \to \infty$ that is, these limits are of the indeterminate form ${1^\infty }$ .

In Section – $3$ , we saw how to evaluate such limits. Writing $f(x)$ as $(1 + h(x))$ reduces this limit to , where $h(x) = f(x) - 1$ .

We will now directly apply this result to evaluate the limits above.

	$\mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{{a^x} + {b^x} + {c^x}}}{3}} \right)^{\dfrac{1}{x}}}$
	$= {e^{\mathop {\lim }\limits_{x \to 0} \dfrac{1}{x} \cdot \left\{ {\dfrac{{{a^x} + {b^x} + {c^x}}}{3} - 1} \right\}}}$
	$= {e^{\mathop {\lim }\limits_{x \to 0} \dfrac{1}{3} \cdot \left\{ {\dfrac{{\left( {{a^x} - 1} \right)}}{x} + \dfrac{{\left( {{b^x} - 1} \right)}}{x} + \dfrac{{\left( {{c^x} - 1} \right)}}{x}} \right\}}}$
	$= {e^{\dfrac{1}{3}\left( {\left( {\ln a} \right) + (\ln b) + (\ln c)} \right)}} = ab{c^{\dfrac{1}{3}}}$

Solution: 7-(b)

	$\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {\dfrac{\pi }{4} + x} \right)} \right)^{\dfrac{1}{x}}}$
	$= {e^{\mathop {\lim }\limits_{x \to 0} \dfrac{1}{x} \cdot \left\{ {\tan \left( {\dfrac{\pi }{4} + x} \right) - 1} \right\}}}$
	$= {e^{\mathop {\lim }\limits_{x \to 0} \left\{ {\dfrac{1}{x} \cdot \dfrac{{2\tan x}}{{1 - \tan x}}} \right\}}}$	$\left( {{\rm{Using}}\tan \left( {\dfrac{\pi }{4} + x} \right) = \dfrac{{1 + \tan x}}{{1 - \tan x}}} \right)$
	$= {e^{\mathop {2\lim }\limits_{x \to 0} \left\{ {\dfrac{{\tan x}}{x} \cdot \dfrac{1}{{1 - \tan x}}} \right\}}}$
	$= {e^2}$

Solution: 7-(c)

	$\mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{\sin \,x}}{x}} \right)^{\dfrac{1}{{{x^2}}}}}$
	$= {e^{\mathop {\lim }\limits_{x \to 0} \dfrac{1}{{{x^2}}} \cdot \left\{ {\dfrac{{\sin x}}{x} - 1} \right\}}}$
	$= {e^{\mathop {\lim }\limits_{x \to 0} \left\{ {\dfrac{{\sin x - x}}{{{x^3}}}} \right\}}}$

In example - $5$ Part – $C$ , we considered the limit in the exponent above.

The value of this limit is therefore ${e^{ - \dfrac{1}{6}}}$

Solution: 7-(d)

	$\mathop {\lim }\limits_{x \to 0} {(\cos x)^{\dfrac{1}{{\sin x}}}}$
	$= {e^{\mathop {\lim }\limits_{x \to 0} \left\{ {\dfrac{1}{{\sin x}}.(\cos x - 1)} \right\}}}$
	$= {e^{\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{ - 2{{\sin }^2}\dfrac{x}{2}}}{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}} \right)}}$
	$= {e^{\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{ - \tan x}}{2}} \right)}} = {e^0} = 1$

Solution: 7-(e)

	$\mathop {\lim }\limits_{x \to 0} {\left( {\cos x + a\,\sin bx} \right)^{\dfrac{1}{x}}}$
	$= {e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{x} \cdot \left\{ {\cos x + a\,\sin bx - 1} \right\}}}$
	$= {e^{\mathop {\lim }\limits_{x \to 0} \left\{ {\dfrac{{\cos x - 1}}{x} + ab\dfrac{{\sin bx}}{{bx}}} \right\}}}$
	$= {e^{\mathop {\lim }\limits_{x \to 0} \left\{ {\dfrac{{ - 2{{\sin }^2}x/2}}{{4{{\left( {x/2} \right)}^2}}} \cdot x + ab\dfrac{{\sin bx}}{{bx}}} \right\}}}$
	$= {e^{\mathop {\lim }\limits_{x \to 0} \left\{ { - \dfrac{1}{2}\dfrac{{{{\sin }^2}\left( {x/2} \right)}}{{{{\left( {x/2} \right)}^2}}} \cdot x + ab\dfrac{{\sin bx}}{{bx}}} \right\}}}$
	$= {e^{ab}}$

Saturday, 9 August 2014

chapter 14 Worked Out Examples 5

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