Methods to Evaluate Limits

EVALUATION OF LIMIT

When the given limit is not in the indeterminate form, limit can be evaluated by directly using the definition of limit or putting a in place of x if x$\to$a. For example$\underset{x\to 3}{\lim }(x+3)=6$Here form is not indeterminate.

When the form of the given function is indeterminate, then our motive will be to get rid of the indeterminate form. After getting rid of the indeterminate form we use the definition of limit or directly put the value. For example $\underset{x\to 2}{\lim }\frac{x^2-4}{x-2}\left(\text{in}\text{determinate}\text{form}\frac{0}{0}\right)$

=$\underset{x\to 2}{\lim }\frac{(x+2)(x-2)}{(x-2)}=\underset{x\to 2}{\lim }(x+2)=4$

METHOD TO EVALUATE THE LIMIT OF AN ALGEBRAIC FUNCTION

The following methods are useful for evaluating limits of algebraic function.

Method of factorisation: If f(x) and g(x) are polynomials such that f(a) = g(a) = 0, then (x – a) is a factor of both f(x) and g(x). Now to solve $\underset{x\to a}{\lim }\frac{f(x)}{g(x)}$, we cancel the common factor (x – a) from both the numerator and denominator, and again put x = a in the given expression. If we get a meaningful number, then the number is the limit of the given expression otherwise we repeat the process till we get rid of indeterminate form.

Method of rationalisation: If in any limit denominator or numerator involve the radical sign, this method is useful.

Example 1: Evaluate $\underset{\text{x}\to \text{a}}{\lim }\frac{\sqrt{x-b}-\sqrt{(a-b)}}{x^2-a^2},a>b$

Solution: $\underset{\text{x}\to \text{a}}{\lim }\frac{\sqrt{x-b}-\sqrt{(a-b)}}{x^2-a^2}.\frac{\sqrt{x-b}+\sqrt{(a-b)}}{\sqrt{x-b}+\sqrt{(a-b)}}$

$\underset{\text{x}\to \text{a}}{\lim }\frac{(x-a)}{\left(x-a\right)\left(x+a\right)\left(\sqrt{x-b}+\sqrt{a-b}\right)}=\frac{1}{4a\sqrt{a-b}}$

Example 2: Evaluate $\underset{\text{x}\to \text{-}\infty }{\lim }\text{x}\left(\text{x+}\sqrt{\text{1+}{\text{x}}^{\text{2}}}\right)$

Common mistakes: How student used to solve this problem.

Step1.$\to$ $\underset{\text{x}\to \text{-}\infty }{\lim }x\left(x+\sqrt{1+x^2}\right).\frac{\sqrt{1+x^2}-x}{\sqrt{1+x^2}-x}$

$\underset{\text{x}\to \text{-}\infty }{\lim }\frac{x}{\sqrt{1+x^2}-x}$

Step2. $\underset{\text{x}\to \text{-}\infty }{\lim }\frac{x}{\sqrt{\frac{1}{x^2}+1}-1}=\infty$

But this is wrong, can you find out where is the mistake.

Solution: $\underset{\text{x}\to \text{-}\infty }{\lim }x\left(x+\sqrt{1+x^2}\right).\frac{\sqrt{1+x^2}-x}{\sqrt{1+x^2}-x}$

Step1. $\underset{\text{x}\to \text{-}\infty }{\lim }\frac{x}{\sqrt{1+x^2}-x}$

Step2. $\underset{\text{x}\to \text{-}\infty }{\lim }\frac{x}{\left|x\right|\sqrt{\frac{1}{x^2}+1}-x}\left(\left|x\right|=-x\text{as}x\to -\infty ,x\text{is}\text{negative}\right)$

$=\underset{\text{x}\to \text{-}\infty }{\lim }\frac{1}{-\sqrt{\frac{1}{x^2}+1}-1}=-\frac{1}{2}$

This is the correct answer

Key concept: Using the concept of method of substitution put

$x=\frac{-1}{t}asx\to -\infty \Rightarrow t\to 0$

$\underset{\text{t}\to 0}{\lim }\frac{\sqrt{1+\frac{1}{t^2}}-\frac{1}{t}}{-t}$

$=\underset{\text{t}\to 0}{\lim }\frac{\left(\sqrt{1+t^2}-1\right)}{-t^2}\times \frac{\sqrt{1+t^2}+1}{\sqrt{1+t^2}+1}$

$=\underset{\text{x}\to \text{-}\infty }{\lim }\frac{1}{-\sqrt{1+t^2}+1}=-\frac{1}{2}$

Method when $x\to \infty$: If given limit is in this form $\underset{\text{x}\to \infty }{\lim }\lim \frac{f\text{(x)}}{\text{g(x)}}$, where f(x) and g(x) are algebraic function in x we divide numerator and denominator by highest power of x in f(x) and g(x)

Example 3: Evaluate $\underset{\text{x}\to \infty }{\lim }\frac{3\sqrt{x}+x^{1/3}-x^{1/4}}{\sqrt{x}+5x^{1/3}+7}$

Solution: Here highest power in f(x) and g(x) is x^1/2. Hence divide numerator and denominator by x^1/2 and then apply the limit.

$\underset{\text{x}\to \infty }{\lim }\frac{3+\frac{1}{x^{1/6}}-\frac{1}{x^{-1/4}}}{1+\frac{5}{x^{1/6}}+\frac{7}{x^{1/2}}}=3$

In this method basically we use the series expansion of sinx, cosx, tanx, log(1+x), a^x, e^x etc to evaluate the limit. Following are some of the frequently used series expansions:

$\bullet$ sin x = $x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots \cdots \cdots$

$\bullet$cos x = $1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots \cdots \cdots$

$\bullet$tan x = $x+\frac{x^3}{3}+\frac{2}{15}{x}^5+\cdots \cdots \cdots$

$\bullet$e^x = $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots \cdots$

$\bullet$a^x = 1 + x.lna + (lna)² $\frac{x^2}{2!}$ +……. a $\in$ R⁺

$\bullet$(1+x)ⁿ = 1 + nx + $\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+\cdots \cdots ,$ n $\in$ R. |x|<1, n is any real number

$\bullet$ln (1+x) = $x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots \cdots ,$ -1 < x $⩽$ 1.

$\bullet$sin^-1 x = $x+\frac{1^2.}{3!}{x}^3+\frac{1^2{.3}^2}{5!}{x}^5+\frac{1^2{.3}^2{.5}^2}{7!}{x}^7+\cdots \cdots \cdots$

$\bullet$tan^-1 x = $x-\frac{x^3}{3}+\frac{x^5}{5}\cdots \cdots$

$\bullet$(1+x)^1/x = $e\left(1-\frac{x}{2}+\frac{11}{24}{x}^2\cdots \cdots \right)$

Example 4: Evaluate $\underset{\text{x}\to 0}{\lim }\frac{\cos x+e^x-2-x}{x^3}$

Solution: $\underset{\text{x}\to 0}{\lim }\frac{\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}....\right)+\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}....\right)-2-x}{x^3}$

$\underset{\text{x}\to 0}{\lim }\text{= lim}\frac{\frac{x^3}{3!}+2.\frac{x^4}{4!}+....}{x^3}$

$\underset{\text{x}\to 0}{\lim }\text{= lim}\frac{1}{3!}+\frac{x}{12}+...(\text{higher}\text{power}\text{of}\text{x})$

= 1/6

APPLICATION OF SOME BASIC LIMITS IN SOLVING THE LIMITS

Following are some basic limits which are used very frequently in solving the limits.

(i) If p(x) is a polynomial, $\underset{\mathbf{x}\to a}{\lim }$p(x) = p(a).

(ii) $\underset{\mathbf{x}\to 0}{\lim }\frac{\sin x}{x}=\underset{\mathbf{x}\to 0}{\lim }\frac{\tan x}{x}=\underset{\mathbf{x}\to 0}{\lim }$ cos x = 1 (where 'x' is in radians)

(iii) $\underset{\mathbf{x}\to 0}{\lim }{\left(1+x\right)}^{1/x}$ = e (iv) $\underset{\mathbf{x}\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^x$ = e

(v) $\underset{\mathbf{x}\to 0}{\lim }\frac{e^x-1}{x}$ = 1 (vi) $\underset{\mathbf{x}\to 0}{\lim }\frac{a^x-1}{x}$= $\ln (a)$, a $\in$ R⁺.

(vii) $\underset{\mathbf{x}\to 0}{\lim }\frac{{(1+x)}^n-1}{x}$ = n (viii) $\underset{\mathbf{x}\to 0}{\lim }\frac{x^m-a^m}{x^n-a^n}=\frac{m}{n}{a}^{m-n}$

(ix) $\underset{x\to 0}{\lim }\lim \frac{\ln \left(1+x\right)}{x}$ = 1 (x) $\underset{\mathbf{x}\to 0}{\lim }\frac{{\log }_a(1+x)}{x}$ = log_a e, a > 0, $\ne$1

(xi) $\underset{\mathbf{x}\to 0}{\lim }\frac{{\sin }^{-1}x}{x}=1=\underset{\mathbf{x}\to 0}{\lim }\frac{{\tan }^{-1}x}{x}$

Now if $\underset{x\to a}{\lim }f(x)=0$ then we can redefine the limits in the following manner

(i) $\underset{\mathbf{x}\to a}{\lim }$$\frac{\sin f(x)}{f(x)}=\underset{x\to a}{\lim }\lim \frac{\tan f(x)}{f(x)}=\underset{x\to a}{\lim }\lim$ cos f(x) = 1(ii) $\underset{\mathbf{x}\to a}{\lim }$$\frac{{\sin }^{-1}f(x)}{f(x)}=\underset{x\to a}{\lim }\frac{{\tan }^{-1}f(x)}{f(x)}$ = 1

(iii) $\underset{\mathbf{x}\to a}{\lim }$$\frac{b^{f(x)}-1}{f(x)}$ = ln b ( b> 0)(iv) $\underset{\mathbf{x}\to a}{\lim }$${\left(1+f(x)\right)}^{1/f(x)}$ = e

Example 5: If f(x) = \left\{ \begin{gathered} \dfrac{{\sin (2k + 3)x}}{{2x}}\,\,\,\,\,\,\,\,\,x < 0 \hfill \\ \,\,\,\,k + 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\, = \,0 \hfill \\ \dfrac{{\tan \left( {3k + 4} \right)x}}{x}\,\,\,\,\,\,\,x\, > \,0 \hfill \\ \end{gathered} \right.

and $\underset{x\to 0}{\lim }f(x)$ exist, then the value of k.

Key concept: Since $\underset{x\to 0}{\lim }f(x)$ exist. That means

$\underset{x\to 0^{+}}{\lim }f(x)=\underset{x\to 0^{-}}{\lim }f(x)$

Solution: $\underset{x\to 0^{+}}{\lim }f(x)=\underset{x\to 0^{+}}{\lim }\frac{\tan (3k+4)x}{x}$

$=\underset{x\to 0^{+}}{\lim }\frac{\tan (3k+4)x}{(3k+4)x}(3k+4)$

$=3k+4\left(\text{as }\underset{\text{x}\to {\text{0}}^{\text{+}}}{\lim }\frac{\tan \left(3k+4\right)x}{x}\right)$

$\underset{x\to 0^{-}}{\lim }f(x)=\underset{x\to 0^{-}}{\lim }\frac{\text{si}n(2k+3)x}{(2k+3)x}.\frac{2k+3}{2}=\frac{2k+3}{2}$

Hence $\text{3k + 4}\text{ =}\frac{\text{2k + 3}}{\text{2}}\Rightarrow k=\frac{-5}{4}$

METHOD OF SUBSTITUTION

To evaluate $\underset{x\to a}{\lim }f\left(x\right)$ by the method of substitution we use different substitutions. For example if we substitute x by a + h, then the given limit becomes $\underset{h\to 0}{\lim }f(a+h)$ (as x$\to$a h$\to$0). Similarly we can use different substitution like substitute x by 2h, 1/h etc. For example, $\underset{x\to a}{\lim }\frac{x^3-a^3}{x^2-a^2}=\underset{h\to 0}{\lim }\frac{{\left(a+h\right)}^3-a^3}{{\left(a+h\right)}^2-a^2}$ (putting x = a + h)

$\underset{h\to 0}{\lim }\frac{h\left(3a^2+3ah+h^2\right)}{2\left(2a+h\right)}=\underset{x\to \infty }{\lim }\frac{3a^2+3ah+h^2}{2a+h}=\frac{3a^2}{2a}=\frac{3}{2}a$

Example 6: Evaluate $\underset{x\to 1}{\lim }=\frac{1-\sqrt{x}}{{\left({\cos }^{-1}x\right)}^2}$

Key concept: First of all substitute cos^–1x = q, use the method of substitution.

Solution: Let cos^–1x = $\theta$, then x = cos$\theta$

Now as x $\to$ 1 $\theta$ $\to$ cos^–1(1), i.e. $\theta$ $\to$ 0

$\underset{x\to 1}{\lim }\frac{1-\sqrt{x}}{{\left({\cos }^{-1}x\right)}^2}=\underset{\theta \to 0}{\lim }\frac{1-\sqrt{\cos \theta }}{{\theta }^{\text{2}}}$

$=\underset{x\to 1}{\lim }\frac{1-\cos \theta }{{\theta }^{\text{2}}\left(1+\sqrt{\cos \theta }\right)}=\underset{\theta \to 0}{\lim }\frac{2{\sin }^2\theta \text{ /2}}{{\theta }^{\text{2}}\left(1+\sqrt{\cos \theta }\right)}$

$=\underset{\theta \to 0}{\lim }\frac{2{\left(\frac{{\sin }^{}\theta \text{ /2}}{\frac{\theta }{\text{2}}}\right)}^2\frac{{\theta }^{\text{2}}}{4}}{{\theta }^{\text{2}}\left(1+\sqrt{\cos \theta }\right)}=\frac{1}{4}$

METHOD FOR CALCULATING THE LIMITS OF THE FORM $\underset{x\to 0}{\lim }f{(x)}^{g(x)}$

If the given limit is in this form then we express the given expression as power of e. For this use the formula $a^x=e^{x{\log }_{e}a}$. Here two cases arises.

Case I: When $\underset{x\to a}{\lim }f(x)=1$

Let $k=\underset{x\to a}{\lim }f{(x)}^{g(x)}$

$k=\underset{x\to a}{\lim }{e}^{g(x)\log f(x)}$

$k={e^{\underset{x\to a}{\lim }}}^{g(x)log\left(f(x)-1\right)}\left(\text{since }\underset{\text{x}\to \text{a}}{\lim }\frac{log(1+f(x)-1)}{(f(x)-1)}\right)$

Hence $\underset{x\to a}{\lim }f{(x)}^{g(x)}$$k={e^{\underset{x\to a}{\lim }}}^{g(x)log\left(f(x)-1\right)}$

Case II: When $\underset{\mathbf{x}\to \mathbf{a}}{\lim }f\left(\mathbf{x}\right)\ne 1$ but f(x) is positive in the neighbourhood of x = a

$\underset{\mathbf{x}\to \mathbf{a}}{\lim }{\left(f\left(x\right)\right)}^{s\left(x\right)}=e^{\underset{x\to a}{\lim }g\left(x\right)\log \left(f\left(x\right)\right)}$

Case III: If f(x) is not throughout positive in the neighbour hood of x = a, then $\underset{\mathbf{x}\to \mathbf{a}}{\lim }{\left(f\left(x\right)\right)}^{g\left(x\right)}$ will not exist. Because in this case function will not be defined in the neighbour hood of x = a

Example 7: Evaluate $\underset{\text{x}\to \text{0}}{\lim }{\text{(1+ax)}}^{\text{b/x}}$

Solution: Clearly given limit is in the form of $\underset{x\to 0}{\lim }f{(x)}^{g(x)}$ where $\underset{x\to a}{\lim }f(x)=1$

Hence $\underset{x\to 0}{\lim }{\text{(1+ ax)}}^{b/x}=e^{\underset{x\to 0}{\lim }\frac{b}{x}(1+ax-1)}$

$\underset{x\to 0}{\lim }{\text{(1+ ax)}}^{b/x}=e^{ab}$

METHOD TO SOLVE THE LIMIT OF THE FORM $\underset{x\to \infty }{\lim }{a}^x$

To solve the limit of the form $\underset{x\to \infty }{\lim }{a}^x$ we use the concept \mathop {{\text{Lim}}}\limits_{{\text{x}} \to \infty } {a^x} = \left\{ \begin{gathered} 0\,\,\,\,\,\,\,\,\,\,\,\,0\,\, \leqslant \,\,a\,\, < \,\,1 \hfill \\ 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = 1 \hfill \\  \infty \,\,\,\,\,\,\,\,\,\,\,\,\,a > 1 \hfill \\ \end{gathered} \right.'

Case I: If $0\le a<1$, then ax will keep on decreasing with on increasing in x.

Hence $\underset{\text{x}\to \infty }{\lim }{a}^x=0$

Case II: If a = 1, in this case given function becomes a constant function. Hence $\underset{\text{x}\to \infty }{\lim }{a}^x=1$.

Case III: If a > 1, then ax will keep on increasing with on increase in the value of x. Hence $\underset{\text{x}\to \infty }{\lim }{a}^x=\infty$

Case IV: If a < 0, then given function will not be defined. Hence $\underset{\text{x}\to \infty }{\lim }{a}^x=\infty$ will not exist.

Misconception: Normally students have a confusion between $\underset{\text{x}\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^x$and$\underset{\text{x}\to \infty }{\lim }{a}^x$when a > 1 and they argue that $\underset{\text{x}\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^x$ should be infinite, because $1+\frac{1}{x}>1\forall x\in R^{+}$

But we have already seen that $\underset{\text{x}\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^x=e$

Actually both the results can not be compared as in the case of $\underset{\text{x}\to \infty }{\lim }{a}^x$, 'a' is a fixed number on the other hand in case of $\underset{\text{x}\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^x,\left(1+\frac{1}{x}\right)$ is clearly a variable number decreasing with increase in x.

Example 8: Evaluate $\underset{\text{x}\to \infty }{\lim }{\left(2^n+3^n+4^n\right)}^{1/n}$

Solution: Given $\underset{\text{x}\to \infty }{\lim }{\left(2^n+3^n+4^n\right)}^{1/n}$ (${\infty }^{\circ }$form)

$\text{=}\underset{\text{n}\to \infty }{\lim }4{\left({\left(\frac{2}{4}\right)}^n+{\left(\frac{3}{4}\right)}^n+1\right)}^{1/n}$

$\text{= }4\left(\text{as }\underset{\text{n}\to \infty }{\lim }{\left(\frac{2}{4}\right)}^n=0,\underset{\text{n}\to \infty }{\lim }{\left(\frac{3}{4}\right)}^n=0\right)$

Example 9: Evaluate $\underset{\text{x}\to \infty }{\lim }\frac{a^x+b^x}{a^x-b^x},\text{where}a>b>1$

Solution: $\underset{\text{x}\to \infty }{\lim }\frac{1+{\left(\frac{b}{a}\right)}^x}{1-{\left(\frac{b}{a}\right)}^x}=1\left(\text{as}\underset{\text{x}\to \infty }{\lim }{(b/a)}^x=0\right)$