Engineering Mathematics V Limits, Continuity and Differentiability

Engineering Mathematics V Limits, Continuity and Differentiability

Mathematics | Limits, Continuity and Differentiability

For a function f(x) the limit of the function at a point x=ais the value the function achieves at a point which is very close to x=a.

Let f(x) be a function defined over some interval containing x=a, except that it
may not be defined at that point.
We say that, L = \lim_{x\to a} f(x) if there is a number \delta for every number \epsilon such that
|f(x)-L| < \epsilon whenever 0<|x-a|<\delta

The concept of limit is explained graphically in the following image –
Definition of limit graphical
As is clear from the above figure, the limit can be approached from either sides of the number line i.e. the limit can be defined in terms of a number less that a or in terms of a number greater than a. Using this criteria there are two types of limits –
Left Hand Limit – If the limit is defined in terms of a number which is less than a then the limit is said to be the left hand limit. It is denoted as x\to a^- which is equivalent to x=a-h where h>0 and h\to 0.
Right Hand Limit – If the limit is defined in terms of a number which is greater than a then the limit is said to be the right hand limit. It is denoted as x\to a^+ which is equivalent to x=a+h where h>0 and h\to 0.

Existence of Limit – The limit of a function f(x) at x=aexists only when its left hand limit and right hand limit exist and are equal i.e.
\lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x)

Some Common Limits –

 \begin{align*} &\bullet\: \lim_{x\to 0} \frac{\sin x}{x} = 1 \hspace{0.5cm}& &\bullet\: \lim_{x\to 0} \cos x = 1 \hspace{0.5cm}& &\bullet\: \lim_{x\to 0} \frac{\tan x}{x} = 1& \\ &\bullet\: \lim_{x\to 0} \frac{1-\cos x}{x} = 0 \hspace{0.5cm}& &\bullet\: \lim_{x\to 0} \frac{\sin x^\circ}{x} = \frac{\pi}{180} \hspace{0.5cm}& &\bullet\: \lim_{x\to a} \frac{x^n - a^n}{x-a} = na^{n-1}& \\ &\bullet\: \lim_{x\to \infty} (1+\frac{k}{x})^{mx} = e^{mk} \hspace{0.5cm}& &\bullet\: \lim_{x\to 0} (1+x)^\frac{1}{x} = e \hspace{0.5cm}& &\bullet\: \lim_{x\to 0} \frac{(a^x-1)}{x} = \ln {a} \hspace{0.5cm}& \\ &\bullet\: \lim_{x\to 0} \frac{e^x-1}{x} = 1 \hspace{0.5cm}& &\bullet\: \lim_{x\to 0} \frac{\ln {(1+x)}}{x} = 1 \hspace{0.5cm}& &\bullet\: \lim_{x\to \infty} x^\frac{1}{x} = 1 \hspace{0.5cm}& \\ \end{align}

L’Hospital Rule –
If the given limit \lim_{x\to a} \frac{f(x)}{g(x)} is of the form \frac{0}{0} or \frac{\infty}{\infty} i.e. both f(x) and g(x) are 0 or both f(x) and g(x) are \infty, then the limit can be solved by L’Hospital Rule.
If the limit is of the form described above, then the L’Hospital Rule says that –
\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f^\prime(x)}{g^\prime(x)}
where f^\prime(x) and g^\prime(x) obtained by differentiating f(x) and g(x).
If after differentitating, the form still exists, then the rule can be applied continuously until the form is changed.

    • Example 1 – Evaluate \lim_{x\to 0} \frac{x\cos{(x)}-\sin{(x)}}{x^2\sin{(x)}}
    • Solution – The limit is of the form \frac{0}{0}, Using L’Hospital Rule and differentiating numerator and denominator=\:\lim_{x\to 0} \frac{\cos{(x)}-x\sin{(x)}-\cos{(x)}}{x^2\cos{(x)} + 2x\sin{(x)}}=\:\lim_{x\to 0} \frac{-\sin{(x)}}{x\cos{(x)} + 2\sin{(x)}}=\:\lim_{x\to 0} \frac{\frac{-\sin{(x)}}{x}}{\cos{(x)} + \frac{2\sin{(x)}}{x}}

      =\:\frac{-1}{1 + 2*1}


    • Example 2 – Evaluate \lim_{x\to 0}\frac{a^{mx}-b^{nx}}{\sin {kx}}
    • Solution – On multiplying and dividing by kx and re-writing the limit we get –
      =\:\lim_{x\to 0}\frac{(a^{mx}-1) - (b^{nx}-1)}{kx}.\frac{kx}{\sin {kx}}=\:\frac{1}{k}\lim_{x\to 0}(\frac{((a^m)^{x}-1)}{x} - \frac{((b^n)^{x}-1)}{x}).\lim_{x\to 0}\frac{kx}{\sin {kx}}=\:\frac{1}{k}(\ln {a^m} -\ln {b^n}).1=\:\frac{1}{k}\ln {\frac{a^m}{b^n}

2. Continuity –

A function is said to be continuous over a range if it’s graph is a single unbroken curve.
A real valued function f(x) is said to be continuous at a point x=x_\circ in the domain if –
\lim_{x\to x_\circ} f(x) exists and is equal to f(x_\circ).
If a function f(x) is continuous at x=x_\circ then-
\lim_{x\to x_\circ ^+} f(x) = \lim_{x\to x_\circ ^-} f(x) = \lim_{x\to x_\circ} f(x)
Functions that are not continuous are said to be discontinuous.

      • Example 1 – For what value of \lambda is the function defined by

         \[ f(x)=\left \{ \begin{tabular}{ll} \lambda(x^2-2),&\:if\:x\leq0 \\ 4x+1,& otherwise\\ \end{tabular} \]

        continuous at x=0?

      • Solution – For the function to be continuous the left hand limit, right hand limit and the value of the function at that point must be equal.
        Value of function at x=0
        f(0) = \lambda * (0-2) = -2\lambda
        Right hand limit-
        =\:\lim_{x\to 0^+} 4x+1=\:1
        RHL equals value of function at 0-
        -2\lambda = 1
        \lambda = \frac{-1}{2}
      • Example 2 – Find all points of discontinuity of the function fdefined by –
      • Solution – The possible points of discontinuity are x=0,1 since the sign of the modulus changes at these points.
        For continuity at x=0,
        =\lim_{x\to 0^-} |x|-|x-1|=\lim_{x\to 0^-} -x-(-(x-1))=\lim_{x\to 0^-} -x+x-1= -1
        =\lim_{x\to 0^+} |x|-|x-1|

        =\lim_{x\to 0^+} x-(-(x-1))

        =\lim_{x\to 0^+} x+x-1

        = -1
        Value of f(x) at x=0,
        f(0) = 0-|0-1| = -1
        Since LHL = RHL = f(0), the function is continuous at x=0

        For continuity at x=1,
        =\lim_{x\to 1^-} |x|-|x-1|

        =\lim_{x\to 1^-} x-(-(x-1))

        =\lim_{x\to 1^-} x+x-1

        = 1
        =\lim_{x\to 1^+} |x|-|x-1|

        =\lim_{x\to 1^+} x-(x-1)

        =\lim_{x\to 1^+} x-x+1

        = 1
        Value of f(x) at x=1,
        f(0) = 1-|1-1| = 1
        Since LHL = RHL = f(1), the function is continuous at x=1
        So, there is no point of discontinuity.

3. Differentiability –

The derivative of a real valued function f(x) wrt x is the function f^\prime(x) and is defined as –
\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}

A function is said to be differentiable if the derivative of the function exists at all points of its domain. For checking the differentiability of a function at point x=c,
\lim_{h\to 0} \frac{f(c+h)-f(c)}{h} must exist.

If a function is differentiable at a point, then it is also continuous at that point.
Note – If a function is continuous at a point does not imply that the function is also differentiable at that point. For example, f(x) = |x| is continuous at x=0 but it is not differentiable at that point.

GATE CS Corner Questions

Practicing the following questions will help you test your knowledge. All questions have been asked in GATE in previous years or in GATE Mock Tests. It is highly recommended that you practice them.

ASSIGNMENT : Engineering Mathematics V Limits, Continuity and Differentiability Assignment MARKS : 10  DURATION : 1 week, 3 days


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