Question

Evaluate the following:
a. $ \underset{x\to 2}{\mathop{\lim }}\,\dfrac{{{x}^{3}}-3{{x}^{2}}+4}{{{x}^{4}}-8{{x}^{2}}+16} $
b. $ \underset{x\to 1}{\mathop{\lim }}\,\left[ \dfrac{2}{1-{{x}^{2}}}+\dfrac{1}{x-1} \right] $
c. $ \underset{x\to 3}{\mathop{\lim }}\,\left[ \dfrac{1}{{{x}^{2}}+4x+3}+\dfrac{1}{{{x}^{2}}+8x+15} \right] $

Answer 1

Hint: After applying the value of limit in the limit expressions if you are getting $ \dfrac{0}{0}or\pm \dfrac{\infty }{\infty } $ form then use L’ Hospital’s rule. In L’ Hospital’s rule, you have to differentiate numerator and denominator and then put the value of limit if it is again in the $ \dfrac{0}{0}or\pm \dfrac{\infty }{\infty } $ then you have to second time differentiate the expression and then apply the limit. You have to do this process until $ \dfrac{0}{0}or\pm \dfrac{\infty }{\infty } $ is eliminated.

Complete step-by-step answer:
 $ \underset{x\to 2}{\mathop{\lim }}\,\dfrac{{{x}^{3}}-3{{x}^{2}}+4}{{{x}^{4}}-8{{x}^{2}}+16} $
 Substituting the value of x = 2 in the above limit expression we get,
 $ \begin{align}
  & \dfrac{{{x}^{3}}-3{{x}^{2}}+4}{{{x}^{4}}-8{{x}^{2}}+16} \\
 & =\dfrac{{{\left( 2 \right)}^{3}}-3{{\left( 2 \right)}^{2}}+4}{{{\left( 2 \right)}^{4}}-8{{\left( 2 \right)}^{2}}+16} \\
 & =\dfrac{8-12+4}{16-32+16} \\
 & =\dfrac{0}{0} \\
\end{align} $
The above limit is in the form of $ \dfrac{0}{0} $ which is an indeterminate form. So, we are going to use L’ Hospital’s rule in which we will differentiate numerator and denominator and then substitute the value of x = 2.
Let us assume f(x) = x3 – 3x2 + 4 and g(x) = x4 – 8x2 + 15 then applying L’ Hospital’s rule we get,
 $ \underset{x\to a}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'(x)}{g'(x)} $ ………………. Eq. (1)
Now, solving f’(x) and g’(x) we get,
 $ f'(x)=3{{x}^{2}}-6x\text{ and }g'(x)=4{{x}^{3}}-16x $
Substituting the above values in eq. (1) we get,
 $ \begin{align}
  & \underset{x\to 2}{\mathop{\lim }}\,\dfrac{3{{x}^{2}}-6x}{4{{x}^{3}}-16x} \\
 & =\underset{x\to 2}{\mathop{\lim }}\,\dfrac{3x-6}{4{{x}^{2}}-16} \\
\end{align} $
Substituting x = 2 in the above limit we get,
 $ \begin{align}
  & \dfrac{3\left( 2 \right)-6}{4\left( 4 \right)-16} \\
 & =\dfrac{0}{0} \\
\end{align} $
As we are getting $ \dfrac{0}{0} $ form so we have to again apply the L’ Hospital’s rule in $ \underset{x\to 2}{\mathop{\lim }}\,\dfrac{3x-6}{4{{x}^{2}}-16} $ then we get,
 $ \underset{x\to 2}{\mathop{\lim }}\,\dfrac{3}{8x} $
Now, we are plugging x = 2 in the above limit we get,
 $ \begin{align}
  & \dfrac{3}{8\left( 2 \right)} \\
 & =\dfrac{3}{16} \\
\end{align} $
Hence, the evaluation of the given limit expression is $ \dfrac{3}{16} $ .

(b) $ \underset{x\to 1}{\mathop{\lim }}\,\left[ \dfrac{2}{1-{{x}^{2}}}+\dfrac{1}{x-1} \right] $
Rewriting the above limit as:
 $ \begin{align}
  & \underset{x\to 1}{\mathop{\lim }}\,\left[ \dfrac{2}{\left( 1+x \right)\left( 1-x \right)}-\dfrac{1}{1-x} \right] \\
 & \underset{x\to 1}{\mathop{\lim }}\,\left[ \dfrac{1}{1-x}\left( \dfrac{2}{1+x}-1 \right) \right] \\
 & \underset{x\to 1}{\mathop{\lim }}\,\left[ \dfrac{1}{1-x}\left( \dfrac{2-1-x}{1+x} \right) \right] \\
 & \underset{x\to 1}{\mathop{\lim }}\,\left[ \dfrac{1}{1-x}\left( \dfrac{1-x}{1+x} \right) \right] \\
\end{align} $
As you can see from the above limit expression that (1 – x) will be cancelled out from numerator and denominator we get,
 $ \underset{x\to 1}{\mathop{\lim }}\,\left[ \dfrac{1}{1+x} \right] $
Now, plugging x = 1 in the above limit expression we get,
 $ \dfrac{1}{2} $
Hence, the value of the limit of the given expression is $ \dfrac{1}{2} $ .

 $ \underset{x\to 3}{\mathop{\lim }}\,\left[ \dfrac{1}{{{x}^{2}}+4x+3}+\dfrac{1}{{{x}^{2}}+8x+15} \right] $ First of all we are making factors of x2 + 4x + 3 and x2 + 8x + 15. $ \begin{align}
  & {{x}^{2}}+4x+3 \\
 & ={{x}^{2}}+x+3x+3 \\
 & =x\left( x+1 \right)+3\left( x+1 \right) \\
 & =\left( x+1 \right)\left( x+3 \right) \\
\end{align} $
 $ \begin{align}
  & {{x}^{2}}+8x+15 \\
 & ={{x}^{2}}+3x+5x+15 \\
 & =x\left( x+3 \right)+5\left( x+3 \right) \\
 & =\left( x+3 \right)\left( x+5 \right) \\
\end{align} $
Now, substituting these factors in the limit expression we get,
 $ \begin{align}
  & \underset{x\to 3}{\mathop{\lim }}\,\left[ \dfrac{1}{\left( x+1 \right)\left( x+3 \right)}+\dfrac{1}{\left( x+3 \right)\left( x+5 \right)} \right] \\
 & \underset{x\to 3}{\mathop{\lim }}\,\left[ \dfrac{1}{x+3}\left( \dfrac{1}{x+1}+\dfrac{1}{x+5} \right) \right] \\
 & \underset{x\to 3}{\mathop{\lim }}\,\left[ \dfrac{1}{x+3}\left( \dfrac{2x+6}{\left( x+1 \right)\left( x+5 \right)} \right) \right] \\
 & \underset{x\to 3}{\mathop{\lim }}\,\left[ \dfrac{2}{x+3}\left( \dfrac{x+3}{\left( x+1 \right)\left( x+5 \right)} \right) \right] \\
 & \underset{x\to 3}{\mathop{\lim }}\,\left[ \dfrac{2}{\left( x+1 \right)\left( x+5 \right)} \right] \\
\end{align} $ Plugging the value of x = 3 in the above limit expression we get, $ \begin{align}
  & \dfrac{2}{\left( 4 \right)\left( 8 \right)} \\
 & =\dfrac{1}{16} \\
\end{align} $
Hence, the value of the given limit expression is $ \dfrac{1}{16} $ .

Note: It is not necessary that limit questions are always solved by L’ Hospital’s rule you can reduce the expression given in question by factorizing or by using basic algebra to not $ \dfrac{0}{0}or\pm \dfrac{\infty }{\infty } $ indeterminate form like we have shown in solving part (b) of the question.