Examples on Evaluating Limits Set-3

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Example – 6

Evaluate the following limits:

\[(a)\quad \mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \cos \,x}}{{{x^2}}}\] \[(b)\quad \mathop {\lim }\limits_{x \to 0} \frac{{x - \ln \left( {1 + x} \right)}}{{{x^2}}}\]
\[(c)\quad \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {1 + x} \right)}^{1/x}} - e}}{x}\] \[(d)\quad  \mathop {\lim }\limits_{x \to 0} \frac{{{{27}^x} - {9^x} - {3^x} + 1}}{{\sqrt 2 - \sqrt {1 + \cos \,x} }}\]

Solution : (a) The limit is of the indeterminate form \(\frac{0}{0}\) , but can be reduced into a combination of two standard limits as follows:

\[\begin{align}& \mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \cos \,x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - 1 + 1 - \cos \,x}}{{{x^2}}} \\&\qquad\qquad\qquad\quad= \mathop {\lim }\limits_{x \to 0} \left\{ {\frac{{{e^{{x^2}}} - 1}}{{{x^2}}} + \frac{{1 - \cos \,x}}{{{x^2}}}} \right\}\\&\qquad\qquad\qquad\quad = \mathop {\lim }\limits_{x \to 0} \left\{ {\frac{{{e^{{x^2}}} - 1}}{{{x^2}}} + \frac{{2\,{{\sin }^2}x/2\,\,}}{{4{{\left( {x/2} \right)}^2}}}} \right\} \\&\qquad\qquad\qquad\quad = 1 + \frac{1}{2} = \frac{3}{2} \\ \end{align} \]

(b) \(\ln \left( {1 + x} \right)\) can be expanded as \(\begin{align}x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - ...\end{align}\)

Hence,

\[\begin{align}& \mathop {\lim }\limits_{x \to 0} \frac{{x - \ln \left( {1 + x} \right)}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \left\{ {\frac{{\frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} + ...}}{{{x^2}}}} \right\} \\ &\qquad\qquad\qquad\qquad= \frac{1}{2} \\ \end{align} \]

(c) The numerator in this limits tends to 0 as \(x \to {\rm{0}}\) because \(\mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{1/x}} = e.\)

Evaluating this limit will require a little artifice in the following manner:

\[\begin{align}&\mathop {\lim }\limits_{x \to 0} \left\{ {\frac{{{{\left( {1 + x} \right)}^{1/x}} - e}}{x}} \right\} = \mathop {\lim }\limits_{x \to 0} \left\{ {\frac{{{e^{\frac{{\ln \left( {1 + x} \right)}}{x}}} - e}}{x}} \right\}\\&= e\mathop {\lim }\limits_{x \to 0} \left\{ {\frac{{{e^{\frac{{\ln \left( {1 + x} \right)}}{x}}} - 1}}{x}} \right\} \text{Taking}\; “e” \;\text{common of the numerator}\end{align}\]

Now as \(x \to {\rm{0,}}\left\{ {\frac{{{\rm{ln}}\left( {{\rm{1 + }}x} \right)}}{x} - 1} \right\} \to 0\) so that the numerator in the limit above is of the form \({e^h} - 1\) Where \(h\,\, \to 0.\)

What should we do now? Multiply and divide by \(h\;\left( {h = \frac{{\ln \left( {1 + x} \right)}}{x} - 1} \right). \)

We get

\[\begin{align} &= e\,\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{e^{\frac{{\ln \left( {1 + x} \right)}}{x} - 1}} - 1}}{{\left\{ {\frac{{\ln \left( {1 + x} \right)}}{x} - 1} \right\}}}} \right] \cdot \left\{ {\frac{{\ln \left( {1 + x} \right)}}{x} - 1} \right\} \cdot \frac{1}{x}\\& = e\,\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{{e^h} - 1}}{h}} \right] \cdot \mathop {\lim }\limits_{x \to 0} \left\{ {\frac{{\ln \left( {1 + x} \right) - x}}{{{x^2}}}} \right\}\end{align}\]

From the previous example, it follows that the second limit has the value \( - \frac{1}{2}\) .

Hence, the overall value for this limit is \( - \frac{e}{2}\)

(d) Factoring the numerator and rationalising the denominator gives.

\[= \mathop {\lim }\limits_{x \to 0} \frac{{\left( {{9^x} - 1} \right)}}{x} \cdot \frac{{\left( {{3^x} - 1} \right)}}{x} \cdot \frac{{\left( {\sqrt 2 + \sqrt {1 + \cos \,x} } \right)}}{{1 - \cos \,x}} \cdot {x^2}\]

(We have multiplied and divided by \({x^2}\) above)

\[\begin{align}&= \mathop {\lim }\limits_{x \to 0} \frac{{\left( {{9^x} - 1} \right)}}{x} \cdot \mathop {\lim }\limits_{x \to 0} \frac{{\left( {{3^x} - 1} \right)}}{x} \cdot \mathop {\lim }\limits_{x \to 0} \left( {\sqrt 2 + \sqrt {1 + \cos x} } \right) \cdot \mathop {\lim }\limits_{x \to 0} \frac{2}{{\frac{{{{\sin }^2}x/2}}{{{{\left( {x/2} \right)}^2}}}}}\\
&= \text{ln }9.\, \text{ln } 3. 2\sqrt 2  . 2\\
&= 8\sqrt 2 {\left( {\ln \,3} \right)^2}\end{align}\]

Example – 7

Evaluate the following limits:

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

Solution: 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 \((1{\rm{ }} + {\rm{ }}h(x))\) reduces this limit to \({e^{\mathop {\lim }\limits_{x \to a} g\left( x \right) \cdot h\left( x \right)}}\) , where \(h\left( x \right){\rm{ }} = {\rm{ }}f\left( x \right){\rm{ }}-{\rm{ }}1.\)

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

 \(\begin{align}{\rm\bf{(a)}} \quad &\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{a^x} + {b^x} + {c^x}}}{3}} \right)^{\frac{1}{x}}}\\&= {e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{x} \cdot \left\{ {\frac{{{a^x} + {b^x} + {c^x}}}{3} - 1} \right\}}}\\&= {e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{3} \cdot \left\{ {\frac{{\left( {{a^x} - 1} \right)}}{x} + \frac{{\left( {{b^x} - 1} \right)}}{x} + \frac{{\left( {{c^x} - 1} \right)}}{x}} \right\}}}\\&= {e^{\frac{1}{3}\left( {\left( {\ln \,a} \right) + \left( {\ln \,b} \right) + \left( {\ln \,c} \right)} \right)}} = ab{c^{1/3}}\end{align}\)

 \(\begin{align}\rm{\bf{(b)}} \quad &\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {\frac{\pi }{4} + x} \right)} \right)^{\frac{1}{x}}}\\&= {e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{x} \cdot \left\{ {\tan \left( {\frac{\pi }{4} + x} \right) - 1} \right\}}}\\&= {e^{\mathop {\lim }\limits_{x \to 0} \left\{ {\frac{1}{x} \cdot \frac{{2\tan \,x}}{{1 - \tan \,x}}} \right\}}}\,\,\,\left( {{\text{Using}}\tan \left( {\frac{\pi }{4} + x} \right) = \frac{{1 + \tan x}}{{1 - \tan x}}} \right)\\& = {e^{\mathop {2\,\lim }\limits_{x \to 0} \left\{ {\frac{{\tan \,x}}{x} \cdot \frac{1}{{1 - \tan \,x}}} \right\}}}\\& = {e^2}\end{align}\)

\(\begin{align}{\rm\bf{(c)}} \quad &\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{\sin \,x}}{x}} \right)^{\frac{1}{{{x^2}}}}}\\&= {e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{{{x^2}}} \cdot \left\{ {\frac{{\sin x}}{x} - 1} \right\}}}\\& = {e^{\mathop {\lim }\limits_{x \to 0} \left\{ {\frac{{\sin x - x}}{{{x^3}}}} \right\}}}\end{align}\)

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

The value of this limit is therefore \({e^{ - 1/6}}\)

 \(\begin{align}{\rm\bf{ (d)}} \quad &\mathop {\lim }\limits_{x \to 0} \,\,{(\cos x)^{\frac{1}{{\sin x}}}}\\&= {e^{\mathop {\lim }\limits_{x \to 0} \,\,\left\{ {\frac{1}{{\sin x}}.(\cos x - 1)} \right\}}}\\&= {e^{\mathop {\lim }\limits_{x \to 0} \,\,\left( {\frac{{ - 2{{\sin }^2}\frac{x}{2}}}{{2\sin \frac{x}{2}\cos \frac{x}{2}}}} \right)}}\\&= {e^{\mathop {\lim }\limits_{x \to 0} \,\,( - \tan x/2)}} = {e^0} = 1\end{align}\)

 \(\begin{align}{\rm\bf{(e)}} \quad &\mathop {\lim }\limits_{x \to 0} {\left( {\cos x + a\,\sin bx} \right)^{\frac{1}{x}}}\\&= {e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{x} \cdot \left\{ {\cos x + a\,\sin bx - 1} \right\}}}\\&= {e^{\mathop {\lim }\limits_{x \to 0} \left\{ {\frac{{\cos x - 1}}{x} + ab\frac{{\sin bx}}{{bx}}} \right\}}}\\& = {e^{\mathop {\lim }\limits_{x \to 0} \left\{ {\frac{{ - 2{{\sin }^2}x/2}}{{4{{\left( {x/2} \right)}^2}}} \cdot x + ab\frac{{\sin bx}}{{bx}}} \right\}}}\\&= {e^{\mathop {\lim }\limits_{x \to 0} \left\{ { - \frac{1}{2}\frac{{{{\sin }^2}\left( {x/2} \right)}}{{{{\left( {x/2} \right)}^2}}} \cdot x + ab\frac{{\sin bx}}{{bx}}} \right\}}}\\&= {e^{ab}}\end{align}\)