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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 1and 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}}
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