Engineering Mathematics Test 4

\(let f(x) = \mathop {\lim }\limits_{\theta \to \infty } \frac{1-cos2\theta}{\theta}\)

\(f(x) = \mathop {\lim }\limits_{\theta \to \infty } \frac{2sin^2\theta}{\theta}\)

Since -1 ≤  sin² θ ≤ 1

Therefore 2sin2 θ  at max can be 2

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

Therefore \( f(x) = \mathop {\lim }\limits_{\theta \to \infty } \frac{1-cos2\theta}{\theta} = 0\)

\(y =\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {x \to 3} \end{array}\frac{{{x^3} - 125}}{{{x^2} - 8x +15}}\)

\(\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {x \to 3} \end{array}\frac{{{x^3} - 125}}{{{x^2} - 8x +15}} = \frac{0}{0}\)

Applying L Hospital’s rule

\(\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {x \to 3} \end{array}\frac{{{3x^2} }}{{{2x} - 8}} = \frac{27}{-2} = \frac{-54}{4}\)

\(f\left( x \right) = \frac{{{x^2} - 3x - 4}}{{{x^2} + 3x - 4}}\)

This function is not defined for:

x² + 3x – 4 = 0

⇒ (x + 4)(x - 1) = 0 i.e.

At x = 1 and x = - 4

\(L=\begin{array}{*{20}{c}}{{\rm{lim}}}\\{h \to 0}\end{array}\frac{{{2^{8cosh}}}}{{8h}}\left[ {{{\sin }^8}\left( {\frac{\pi }{6} + h} \right) - {{\sin }^8}\frac{\pi }{6}} \right]\)

put h = 0

\(L = \;\frac{{{2^{8cos0}}}}{{8 \times 0}}\left[ {{{\sin }^8}\left( {\frac{\pi }{6} + 0} \right) - {{\sin }^8}\frac{\pi }{6}} \right]\)

\(L = \frac{0}{0}\)

It is of the form

Apply L Hospital

\(L\; = \begin{array}{*{20}{c}}{{\rm{lim}}}\\{h \to 0}\end{array}\frac{{{2^{8cosh}}\left[ {8{{\sin }^7}\left( {\frac{\pi }{6} + h} \right){\rm{cos}}\left( {\frac{{\rm{\pi }}}{6}{\rm{\;}}} \right) - 0} \right] + \left[ {{{\sin }^8}\left( {\frac{\pi }{6} + h} \right) - {{\sin }^8}\frac{\pi }{6}} \right]\;{2^{8cosh\;}}\left( { - sinh} \right)}}{{8 \times 1}}\)

\(L\; = \frac{{{2^{8cos0}}\left[ {8{{\sin }^7}\left( {\frac{\pi }{6}} \right){\rm{cos}}\left( {\frac{{\rm{\pi }}}{6}{\rm{\;}}} \right)]{\rm{\;}}} \right] + 0}}{8}\)

\(L\; = \;\;\frac{{{2^8}\left[ {8 \times {{\left( {\frac{1}{2}} \right)}^7} \times \left( {\frac{{\surd 3}}{2}} \right)} \right]}}{8}\)

using L Hospital's

\(\mathop {\lim }\limits_{x \to 0}\frac{{128^xlog128 - 4^xlog4}}{{a^xloga}} =5\)

\( \frac{128^0{log128 - 4^0log4}}{{a^0loga}} =5\)

\( \frac{{log128 - log4}}{{loga}} = 5\)

\(5loga = \log \left( {\frac{{128}}{4}} \right)\)

\(5loga = \log \left( 2^{5} \right)\)

\(5loga = 5\log \left( 2 \right)\)

∴ a = 2

\(f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \frac{{x + {x^2} + {x^3} + {x^4} + \ldots + {x^n} - n}}{{x - 1}}\)

\(f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \frac{{x + {x^2} + {x^3} + {x^4} + \ldots + {x^n} - (1+1+1+1+..n)}}{{x - 1}}\)

\(f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \left( {\frac{{x - 1}}{{x - 1}} + \frac{{{x^2} - 1}}{{x - 1}} + \frac{{{x^3} - 1}}{{x - 1}} + \frac{{{x^4} - 1}}{{x - 1}} + \ldots } \right)\)

\(f\left( x \right) = \left( {1 + 2 + 3 + 4 + \ldots } \right)\)

\(f\left( x \right) = \frac{{n\left( {n + 1} \right)}}{2}\)

Concept:

A function f(x) is continuous at x = a if,

Left limit = Right limit = Function value = Real and finite

A function is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

Calculation:

Given:

f(x) = |x|

|x| = x for x ≥ 0

|x|= -x for x < 0

At x = 0

Left limit = 0, Right limit = 0, F(0) = 0

As

Left limit = Right limit = Function value = 0

∴ |X| is continuous at x = 0.

Now

Left derivative (at x = 0) = -1

Right derivative (at x = 0) = 1

Left derivative ≠ Right derivative

∴ |x| is not differentiable at x = 0

Concept:

If \(\begin{array}{*{20}{c}} {lim}\\ {x \to a} \end{array}\;f\left( x \right) = 1\;\& \;\begin{array}{*{20}{c}} {lim}\\ {x \to a} \end{array}\;g\left( x \right) = \infty\)

i.e. 1^∞ form

then, \(\begin{array}{*{20}{c}} {lim}\\ {x \to a} \end{array}{\left[ {f\left( x \right)} \right]^{g\left( x \right)}}\)

\({e^{\begin{array}{*{20}{c}} {lim}\\ {x \to a} \end{array}}}\;\phi \left( x \right)\left[ {f\left( x \right) - 1} \right]a\)

\(\begin{array}{*{20}{c}} {lim}\\ {x \to 0} \end{array}\;f\left( x \right) = 1 + {\tan ^2}\sqrt x = 1\)

\(\begin{array}{*{20}{c}} {lim}\\ {x \to 0} \end{array}\;g\left( x \right) = \frac{1}{{2x}} = \infty\)

\({e^{\begin{array}{*{20}{c}} {lim}\\ {x \to 0} \end{array}}}\frac{1}{{2x}}\left[ {1 + {{\tan }^2}\sqrt x - 1} \right]\)

\(p = {e^{\mathop {\lim }\limits_{x \to 0} {{\left[ {\frac{{\tan \left( {\sqrt x } \right)}}{{\sqrt x }}} \right]}^2}}}\)

p = e^1/2

\(\ln p = \frac{1}{2}\)

\(let\;y = \mathop {{\rm{lim}}}\limits_{x \to 0\;} \frac{{{{({5^x} - 1)}^4}}}{{({7^x} - 1).sinx.\;{\rm{log}}\left( {1 + x} \right).tanx}}\)

Since x → 0 ∴ x⁴ → 0

Divide number and denominator by x⁴

\(\mathop {{\rm{lim}}}\limits_{x \to 0\;} \frac{{\frac{{{{\left( {{5^x} - 1} \right)}^4}}}{{{x^4}}}}}{{\;\frac{{({7^x} - 1).sinx.\;{\rm{log}}\left( {1 + x} \right).tanx}}{{{x^4}}}\;}}\)

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{{a^x} - 1}}{x} = lo{g_e}a\)

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{sinx}}{x} = 1\)

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{tanx}}{x} = 1\)

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{1 + logx}}{x} = 1\)

\(y = \frac{{{{(\log 5)}^4}\;}}{{log7}}\)