Limits MCQ Quiz in தமிழ் - Objective Question with Answer for Limits - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Limit:

The number L is called the limit of function f(x) as x→a if and only if, for every ε > 0 there exists δ > 0 such that:

|f(x) - L| < ϵ 

Whenever 0 < |x - a| < δ 

Calculation:

Given that:

xn = 0.5xn−1 + 1 

\(x_n = \frac{x_{n-1}}{2}+1\)

x₁ = 1 (x₀ = 0 given)

\(x_2 = \frac{x_1}{2} + 1 = \frac{1}{2} + 1 = \frac{3}{2}\)

\(x_3 = \frac{x_2}{2} + 1 = \frac{3}{2 \times 2} + 1 = \frac{7}{4}\)

\(x_4 = \frac{x_3}{2} + 1 = \frac{7}{4 \times 2} + 1 = \frac{15}{8}\)

\(x_5 = \frac{x_4}{2} + 1 = \frac{15}{8 \times 2} + 1 = \frac{31}{16}\)

The sequence is:

\(= 1, \frac{3}{2}, \frac{7}{2^2}, \frac{15}{2^3}, \frac{31}{2^4}, .. , \frac{2^n -1}{2^{n-1}},..\)

\(\mathop {\lim }\limits_{n \to \infty } {x_n} = \mathop {\lim }\limits_{n \to \infty } \frac{{{2^n} - 1}}{{{2^{n - 1}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{2^n} - 1}}{{{2^n}}}2\)

\(\mathop {\lim }\limits_{n \to \infty } {x_n} = \mathop {\lim }\limits_{n \to \infty } 2\left( {1 - \frac{1}{{{2^n}}}} \right) = 2\)

Explanation:

\(\rm p = \displaystyle\lim_{x \rightarrow π} \left( \frac{x^2 + α x + 2 π^2}{x - π + 2 \sin x} \right)\)

substitue x = π

\(\rm p = \left( \frac{π^2 + α π + 2 π^2}{π - π + 2 \sin π} \right )\), sin π = 0

We can see that denominator becomes zero.

Thus for p to have finite value numerator is also zero.

π² + απ + 2π² = 0

α = - 3π

Now,  \(\rm p = \left( \frac{\pi ^2 + α π + 2 π^2}{\pi - π + 2 \sin \pi} \right ) = \frac{0}{0} \)form

Using L'Hospital's rule, we get

\(\rm p = \displaystyle\lim_{x \rightarrow π} \left( \frac{2x\ +\ α }{1\ +\ 2 \cos x} \right ) \)

substituting x = π, α = - 3π

\(\rm p = \left( \frac{2π\ -\ 3π }{1\ +\ 2 \cos π} \right ) \), cosπ = -1

p = \(\frac{-π}{-1}\)

p = π 

Concept:

L’ Hospital’s Rule:

\(\mathop {\lim }\limits_{x \to a} \left\{ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right\} = \mathop {\lim }\limits_{x \to a} \left\{ {\frac{{f'\left( x \right)}}{{g'\left( x \right)}}} \right\}\)

This rule is only applicable for \(\frac{0}{0}\) and \(\frac{\infty }{\infty }\) indeterminate forms.

Calculation

As 

\(\mathop {\lim }\limits_{x \to \infty } \frac{{xln\;x}}{{1 + {x^2}}} = \frac{\infty }{\infty }\)

Which is an indeterminate form.

Using L hospitality rule for \(\mathop {\lim }\limits_{x \to \infty } \frac{{xln\;x}}{{1 + {x^2}}}\)

We get

\(\mathop {\lim }\limits_{x \to \infty } \frac{{x \times \frac{1}{x} + lnx}}{{2x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{1 + lnx}}{{2x}} = \frac{\infty }{\infty }\;Indeterminate\;form\)

Again applying the L hospitality rule, we get 

\(\mathop {\lim }\limits_{x \to \infty } \frac{{\frac{1}{x}}}{2} = \frac{{\frac{1}{\infty }}}{2} = 0\)

∴ Value of 

\(\mathop {\lim }\limits_{x \to \infty } \frac{{xln\;x}}{{1 + {x^2}}} = 0\)

\(let\;y = {x^{\frac{1}{x}}}\)

Taking log_e on both sides

\({\log _{\rm{e}}}{\rm{y}} = \frac{1}{{\rm{x}}}{\log _{\rm{e}}}{\rm{x}}\)

\({\rm{Taking\;}}\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to \infty } {\rm{\;on\;both\;sides\;}}\)

\(\mathop {{\rm{lim}}}\limits_{x \to \infty } {\log _{\rm{e}}}{\rm{y}} = \mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{1}{{\rm{x}}}{\log _{\rm{e}}}{\rm{x}}\)

Substituting x = ∞

It is of the form \(\frac{\infty}{\infty}\) 

Differentiating w.r.t x

\(\mathop {{\rm{lim}}}\limits_{x \to \infty } {\log _{\rm{e}}}{\rm{y}} = \mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{1}{{\rm{x}}}\)

Substituting x = ∞

\({\log _{\rm{e}}}{\rm{y}} = 0\;\;\;\;\;\;\;\left( {\frac{1}{\infty } = 0} \right)\)

∴ y = e⁰

\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \left( {1 + x + \frac{{{x^2}}}{2}} \right)}}{{{x^3}}} = \frac{0}{0}\)Not defined 

Using L – Hospital’s rule,

1^st derivation

\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\frac{d}{{dx}}\left[ {{e^x} - \left( {1 + x + \frac{{{x^2}}}{2}} \right)} \right]}}{{\frac{d}{{dx}}\left( {{x^3}} \right)}} = {\lim _{{\rm{x}} \to 0}}\frac{{{e^x} - \left( {0 + 1 + \frac{{2x}}{2}} \right)}}{{3{x^2}}} = \frac{{1 - \left( {0 + 1 + 0} \right)}}{0} = \frac{0}{0}\)Not defined

2^nd derivation

\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{{\rm{e}}^{\rm{x}}} - \left( {0 + 0 + 1} \right)}}{{6{\rm{x}}}} = \frac{0}{0}\)Not defined

3^rd derivation

\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{e^x} - 0}}{6} = \frac{1}{6}\)

Concept: 

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I.  \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}\)

II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule as:

\(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)

Calculation:

Given​​

\(\displaystyle\lim_{x \rightarrow 0} \left(\dfrac{1 - \cos x}{x^2}\right) = \left( {\frac{0}{0}} \right)\)

Applying L’ Hospital  rule:

\(\displaystyle\lim_{x \rightarrow 0} \left(\dfrac{1 - \cos x}{x^2}\right) =\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{\frac{d}{dx}({1 - \cos x})}{{\frac{d}{dx}(x^2})} =\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{sin\;x {\rm{}}}}{{ {2x\rm}}} \)

\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{sin\;x {\rm{}}}}{{ {2x\rm}}} = \left( {\frac{0}{0}} \right){\rm{form}}\)

Once again by L’ Hospital rule,

\({\rm{}}\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{cos\; {\rm{x}}}}{{{\rm{2}}}} = \frac{1}{2}\)

Concept:

\(\begin{array}{*{20}{c}} {\lim}\\ {x \to 0} \end{array}\frac{{\sin x}}{x} = 1\)

Calculation:

Given:

\(\begin{array}{*{20}{c}} {\lim}\\ {x \to 0} \end{array}\frac{{\sin \left( {\pi {{\cos }^2}x} \right)}}{{{x^2}}}\)

cos² x = 1 – sin² x

\(\begin{array}{*{20}{c}} {\lim}\\ {x \to 0} \end{array}\frac{{\sin \left( {\pi {{\cos }^2}x} \right)}}{{{x^2}}} = \begin{array}{*{20}{c}} {\lim}\\ {x \to 0} \end{array}\frac{{\sin \left\{ {\pi \left( {1 - {{\sin }^2}x} \right)} \right\}}}{{{x^2}}}\)

\(= \begin{array}{*{20}{c}} {\lim}\\ {x \to 0} \end{array}\frac{{\sin \left( {\pi - \pi \;si{n^2}x} \right)}}{{{x^2}}}\)

\(= \begin{array}{*{20}{c}} {\lim}\\ {x \to 0} \end{array}\frac{{\sin \left( {\pi si{n^2}x} \right)}}{{{x^2}}}\)

Multiply and divide by π sin²x

\( = \begin{array}{*{20}{c}} {\lim}\\ {x \to 0} \end{array}\frac{{\sin \left( {\pi si{n^2}x} \right)}}{{\pi {{\sin }^2}x}} \times \frac{{\pi {{\sin }^2}x}}{{{x^2}}}\)

\(Since~lim_{x\rightarrow 0}\frac{sinx}{x}=1\), the above expression gives:

Concept:

L’ Hospital’s Rule:

\(\mathop {\lim }\limits_{x \to a} \left\{ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right\} = \mathop {\lim }\limits_{x \to a} \left\{ {\frac{{f'\left( x \right)}}{{g'\left( x \right)}}} \right\}\)

This rule is only applicable for \(\frac{0}{0}\) and \(\frac{\infty }{\infty }\) indeterminate forms.

Explanation:

\(\mathop {\lim }\limits_{x \to 1} (\frac{1}{1n x}-\frac{1}{x-1})\)

\(\mathop {\lim }\limits_{x \to 1} \left[ {\frac{{\left( {x - 1} \right) - \ln x}}{{\ln x\;\left( {x - 1} \right)}}} \right]\) which is in the form of 0/0

So, applying L- Hospital rule twice,

\(\mathop {\lim }\limits_{x \to 1} \left[ {\frac{{1 - \frac{1}{x}}}{{\log x + \left( {x - 1} \right)\left( {\frac{1}{x}} \right)}}} \right] \)

then,

\(\mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{{{x^2}}}}}{{\frac{1}{{{x^2}}} + \frac{1}{x}}}\)

= \(\frac{1}{{1 + 1}} = \frac{1}{2} = 0.5\)

\(\mathop {\lim }\limits_{x \to \infty } \left( {ax + \left( {\frac{{7 - \sqrt 3 {x^2}}}{{3 - x}}} \right)} \right) = b\;\)

\(ax + \frac{{7 - \sqrt 3 {x^2}}}{{3 - x}}\)

\( = \frac{{ax\left( {3 - x} \right) + 7 - \sqrt 3 \;{x^2}}}{{3 - x}}\)

\( = \frac{{3ax - a{x^2} + 7 - \sqrt 3 \;{x^2}}}{{3 - x}}\)

\( = \frac{{ - \sqrt 3 \;{x^2} - a{x^2} + 3ax + 7}}{{3 - x}}\)

\(\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}\left( { - \sqrt 3 - a} \right) + x\left( {3a + \frac{7}{x}} \right)}}{{x\left( {\frac{3}{x} - 1} \right)}}\)

∴ For the limit to be finite, the x² term in the numerator should be 0.
\(\therefore \; - \sqrt 3 - a = 0\)

\(a = - \sqrt 3 \)

Now,

\(\mathop {\lim }\limits_{x \to \infty } \frac{{x\left( { - 3\sqrt 3 + \frac{7}{x}} \right)}}{{x\left( {\frac{3}{x} - 1} \right)}}\)

\( = \frac{{ - 3\sqrt 3 }}{{ - 1}} = 3\sqrt 3 \)

Concept:

A function f(x) is said to be continuous at a point x = a, in its domain if,

\(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{a}} \right)\) exists or its graph is a single unbroken curve.

f(x) is Continuous at x = a

\(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right)\)

Calculation:

Given:

\(f\left( x \right) = \frac{{\left| x \right|}}{x}\)

\(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \;\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\left| x \right|}}{x} = 1\)

\(for\;x \to {0^ + },\;\left| x \right| = x\)

\(\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \;\mathop {\lim }\limits_{x \to {0^ - }} \frac{{\left| x \right|}}{x} = \frac{{\left| { - x} \right|}}{{ - x}} = \frac{x}{{ - x}} = \; - 1\)

\(\)Left hand limit and right hand limit are not equal 

∴ \(\mathop {\lim }\limits_{x \to 0} f\left( x \right)\;\)does not exist.

∴ All given options are true.