log(3+x) -log(3- x)
(i) lim
x 0
X
Answers
Answer:
The limit equation is given by:
\lim_{x \to 0} \left( \frac{sin(2x)}{x^3} + a + \frac{b}{x^2} \right) = 0lim
x→0
( x 3sin(2x)
+a+ x 2b)=0
Separating the constant aa from the limit, we have:
\lim_{x \to 0} \left( \frac{\sin (2x)}{x^3} + \frac{b}{x^2} \right) + a = 0lim
x→0
( x 3
sin(2x)
+ x 2b)+a=0
\Rightarrow \lim_{x \to 0} \left( \frac{ \sin (2x) + bx }{x^3} \right) + a = 0⇒lim
x→0
( x 3
sin(2x)+bx)+a=0
After substituting the limit value x = 0x=0 into the limit function, we can see that we have \frac{0}{0} 00
form is exist. So we will apply L'Hospital's Rule to this limit function.
Now:
We know that this rule states that if such indeterminate form \frac{0}{0} 00
exists then we will take the derivatives of the numerator and denominator and substitute the limit value. This process will apply again and again until we get a finite value.
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Applying L'Hospital's Rule to the limit function, we have:
\lim_{x \to 0} \left( \frac{2 \cos (2x) + b }{3x^2} \right) + a = 0lim
x→0
( 3x22cos(2x)+b )+a=0
Substituting x = 0x=0 into the limit function, we have:
\Rightarrow \left( \frac{2 + b}{3(0^2)} \right) + a \hspace{1 cm} \left[ \because \cos (0) = 1 \right]⇒(
3(0 2)2+b )+a[∵cos(0)=1
After substituting x = 0x=0 into the limit function, we get \infty∞ . But this result is not possible because the right hand side of the limit equation is zero. So, we have to find a value of bb so that b + 2 = 0 \, or \, b = - 2b+2=0orb=−2
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Now:
Substituting b = - 2b=−2 into the limit function we have:
\lim_{x \to 0} \left( \frac{2 \cos(2x) - 2}{3x^2} \right) + a = 0lim
x→0
( 3x 2
2cos(2x)−2)+a=0
\Rightarrow \frac{2}{3} \lim_{x \to 0} \left( \frac{\cos (2x) - 1}{x^2} \right) + a = 0⇒ 32
lim
x→0 ( x 2
cos(2x)−1 )+a=0
\Rightarrow \frac{2}{3} \lim_{x \to 0} \left( \frac{ - 2 \sin (2x) - 0}{2x} \right) + a = 0⇒ 32
lim
x→0
( 2x−2sin(2x)−0
)+a=0 \hspace{1 cm} [ \because \frac{0}{0}∵ 00
form, so we used L'Hospital's Rule ]
\Rightarrow - \frac{2}{3} \lim_{x \to 0} \left( \frac{ \sin (2x)}{x} \right) + a = 0⇒−32
lim
x→0
( xsin(2x) )+a=0
\Rightarrow - \frac{2}{3} \lim_{x \to 0} \left( \frac{2 \cos(2x)}{1} \right) + a = 0 \hspace{1 cm}⇒− 32 lim
x→0 ( 1 )
2cos(2x) )+a=0 [ \because \frac{0}{0}∵ 00
form, so we used L'Hospital's Rule ]
\Rightarrow - \frac{2}{3} \left( 2 \cos(0) \right) + a = 0 \hspace{1 cm}⇒− 32
(2cos(0))+a=0 [ Substitute x = 0x=0 into the limit function ]
\Rightarrow - \frac{4}{3} + a = 0 \hspace{1 cm} \left[ \because \cos(0) = 1 \right]⇒− 34+a=0[∵cos(0)=1]
\Rightarrow a = \frac{4}{3}⇒a= 34
Hence:
The values of a\, \, and \,\, baandb are \frac{4}{3} \, and \, -2
34and−2 respectively.