How To Solve Limit Problems

Limit problems’ first post by Online IB Tutors was all about some basic as well as advanced concepts of limits problems. Here we shall discuss different methods to solve limit questions.

How To Solve Limit Problems

Based on the type of function, we can divide all our work into sections-:

Algebraic Limit Problems- Problems that involve algebraic functions are called algebraic limits. They can be further divided into the following sections:-

Direct Substitution Method –Suppose we have to find. L = {\lim }\limits_{x \to a} f(x) we can directly substitute the value of the limit of the variable (i.e replace x=a) in the expression.

► If f(a) is finite then L=f(a)

► If f(a) is undefined then L doesn’t exist

► If f(a) is indeterminate  then this method fails

<img src="limit.png" alt="limit">

Example-1:- Find value of   {\lim }\limits_{x \to 2} (x²-5x+6)

 

Ans:   {\lim }\limits_{x \to 2}  (x²-5x+6)

=2²-5.2+6

=4-10+6

=0

Factorization Method – Suppose we need to find   L = {\lim }\limits_{x \to a} \frac{{P(x)}}{{Q(x)}}   where P(x) and Q(x) are polynomials, then we factorize both P(x) and Q(x) in their lowest form. Then we simplify the given expression as much as possible. After all this, we put the limit. We try to ensure that we don’t get zero in the denominator.

we can use these tricks-

► {x^n} - {a^n} = (x - a)({x^{n - 1}} + {x^{n - 2}}{a^1} + {x^{n - 2}}{a^1} + {x^{n - 3}}{a^2} + {x^{n - 4}}{a^3}............... + {a^{n - 1}})

where n can be even or odd positive integer

► {x^n} + {a^n} = (x + a)({x^{n - 1}} - {x^{n - 2}}{a^1} + {x^{n - 2}}{a^1} - {x^{n - 3}}{a^2} + {x^{n - 4}}{a^3}............... + {( - 1)^{n - 1}}{a^{n - 1}})

where n is an odd positive integer. This formula is not applicable when n is even

Sometimes, we can directly use  below formula to evaluate the limit

 {\lim }\limits_{x \to a} \frac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}

If the degree of the numerator is more than or equal to the degree of the denominator, then we should divide.

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Example-2:- Evaluate   {\lim }\limits_{x \to 2} \left( {\frac{{{x^3} - 2x - 4}}{{{x^2} - 3x + 2}}} \right)

 

Ans:-         {\lim }\limits_{x \to 2} \left( {\frac{{{x^3} - 2x - 4}}{{{x^2} - 3x + 2}}} \right)

 

 = {\lim }\limits_{x \to 2} \left( {\frac{{(x - 2)({x^2} + 2x + 2)}}{{{x^2} - 3x + 2}}} \right)

 

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

 

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

 

 = \frac{{4 + 4 + 2}}{{2 - 1}}

 

= 10

 

 

Rationalization Method:- If we ever get 0/0 form in the problems involving square roots, then there must be a common factor in both numerator and denominator which must be canceled out to get a meaningful form. To cancel this common factor, we rationalize the denominator or numerator or both.

Example-3:-Evaluate   {\lim }\limits_{x \to 3} \frac{{\sqrt {3x + 7} - 4}}{{\sqrt {x + } 1 - 2}}

 

Ans-              {\lim }\limits_{x \to 3} \frac{{\sqrt {3x + 7} - 4}}{{\sqrt {x + } 1 - 2}}

Rationalizing the denominator and the numerator both

 

=     {\lim }\limits_{x \to 3} \frac{{3x + 7 - 16}}{{x + 1 - 4}}\left( {\frac{{\sqrt {x + 1} + 2}}{{\sqrt {3x + 7} + 4}}} \right)

 

=   {\lim }\limits_{x \to 3} \frac{{3x - 9}}{{x - 3}}\left( {\frac{{\sqrt {x + 1} + 2}}{{\sqrt {3x + 7} + 4}}} \right)

 

=    {\lim }\limits_{x \to 3} \frac{{3(x - 3)}}{{x - 3}}\left( {\frac{{\sqrt {x + 1} + 2}}{{\sqrt {3x + 7} + 4}}} \right)

 

=  3.  {\lim }\limits_{x \to 3} \left( {\frac{{\sqrt {x + 1} + 2}}{{\sqrt {3x + 7} + 4}}} \right)

 

= 3.  \left( {\frac{{\sqrt {3 + 1} + 2}}{{\sqrt {3 \times 3 + 7} + 4}}} \right)

 

\frac{{3 \times 4}}{8} = \frac{3}{2}

Solving of not-defined type Limit problems

 If we are given a problem with  {\lim }\limits_{x \to \infty }  we first of all note the highest st power of x in the whole question. After this, we divide both numerator and denominator by that power. This will convert both numerator and denominator into 1/x form. After this, we can use the following formula

 {\lim }\limits_{x \to \infty } \frac{1}{{{x^n}}} = 0

we can use the below result to solve most problems of this category

                            {\lim }\limits_{x \to \infty } \frac{{a{}_0{x^m} + a{}_1{x^{m - 1}} + .......... + a{}_m}}{{b{}_0{x^n} + b{}_1{x^{n - 1}} + ........... + b{}_m}}

► If m=n then our answer is   {\frac{{a{}_0}}{{b{}_0}}}

► If m<n then our answer is  0

► If m>n then our answer is  not defined

 

 

Based On Logarithmic and Exponential Functions:- There are several short cut tricks that we can use to solve these types of problems. A few of them are as follows-

Also READ this-  Continuity of functions | Learn Maths Online

1.    {\lim }\limits_{x \to 0} \frac{{\log {}_e(1 + x)}}{x} = 1

2.  {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x} = 1

3.   {\lim }\limits_{x \to 0} \log {}_e{\{ 1 + f(x)\} ^{\frac{1}{{f(x)}}}} = e

4.  {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = 1

5. If  {\lim }\limits_{x \to a} f(x) = 1 and   {\lim }\limits_{x \to a} g(x) = \infty    then    {\lim }\limits_{x \to a} f(x){}^{g(x)} = {e^{ {\lim }\limits_{x \to a} g(x)[f(x) - 1]}}

6.   {\lim }\limits_{x \to \infty } {\left( {\frac{{x + a}}{{x + b}}} \right)^{cx + d}} = {e^{(a - b)c}}

7.  {\lim }\limits_{x \to \infty } {\left( {\frac{{x{}^2 + ax + b}}{{x{}^2 + cx + d}}} \right)^{px + q}} = {e^{p(a - c)}}

8.  {\lim }\limits_{x \to a} \frac{{{b^{f(x)}} - 1}}{{f(x)}} = \ln b

Example-3:- Evaluate     {\lim }\limits_{x \to 0} \frac{{{e^{\tan x}} - {e^x}}}{{(\tan x - x)}}

 

Ans:-            {\lim }\limits_{x \to 0} \frac{{{e^{\tan x}} - {e^x}}}{{(\tan x - x)}}

 

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

 

=    {\lim }\limits_{x \to 0} {e^x} {\lim }\limits_{x \to 0} \frac{{({e^{\tan x - x}} - 1)}}{{(\tan x - x)}}           (by rule viii)

 

=  {e^0} \times 1

 

=1

 

Problems Based on Series:-

(i)  ex =1+x/1!+x3/3!+x4/4!……\infty

 

(ii)  ax=1+(xloga)/1!+ (xloga)2/2!+ (xloga)3/3!+ (xloga)4/4!+……….where a > 0

 

(iii)   ln(1-x)=x-x2/2+x3/3-x4/4……….    where -1 < x  1

 

(iv)  ln(1-x)=-x-x2/2-x3/3-x4/4……….     where  -1 x < 1

 

(v )  \sin x = x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - \frac{{{x^7}}}{{7!}}.......

 

(vi) \cos x = 1 - \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} - \frac{{{x^6}}}{{6!}}.......

 

(v)  \tan x = x + \frac{{{x^3}}}{3} + \frac{{2{x^5}}}{{5!}} - ..........

 

(vi)  {\sin ^{ - 1}}x = x + \frac{{{1^2}.x{}^3}}{{3!}} + \frac{{{1^2}.3{}^2x{}^5}}{{5!}} + \frac{{{1^2}{{.3}^2}{{.5}^2}x{}^3}}{{7!}}..........

(vii{\tan ^{ - 1}}x = x - \frac{{x{}^3}}{3} + \frac{{x{}^5}}{5}..........

 

(viii) {(1 + x)^n} = 1 + nx + \frac{{n(n - 1)}}{{2!}}{x^2} + \frac{{n(n - 1)(n - 2)}}{{3!}}{x^3}.......

Examples:-   Find Value of     {\lim }\limits_{x \to 2} \frac{{{3^x} + {3^{3 - x}} - 12}}{{{3^{3 - x}} - {3^{\frac{x}{2}}}}}

 

Ans:-              {\lim }\limits_{x \to 2} \frac{{{3^x} + {3^{3 - x}} - 12}}{{{3^{3 - x}} - {3^{\frac{x}{2}}}}}

 {\lim }\limits_{x \to 2} \frac{{{3^x} + \frac{{{3^3}}}{{{3^x}}} - 12}}{{\frac{{{3^3}}}{{{3^x}}} - {3^{\frac{x}{2}}}}}

 

 {\lim }\limits_{x \to 2} \frac{{{3^{2x}} - {{12.3}^x} + 27}}{{{3^3} - {3^{\frac{{3x}}{2}}}}}

=   {\lim }\limits_{x \to 2} \frac{{({3^x} - 9)({3^x} - 3)}}{{({3^{x/2}} - 3)({3^x} + {{33}^{x/2}} + 9)}}

 

                                                =    {\lim }\limits_{x \to 2} \frac{{({3^{x/2}} + 3)({3^x} - 3)}}{{({3^x} + {{33}^{x/2}} + 9)}}

 

                                               =    \frac{{ - 6 \times 6}}{{9 + 3.3 + 9}}

 

=    \frac{{ - 36}}{{27}}

 

=    \frac{{ - 4}}{3}

Limits Based on Trigonometric Functions:-

 {\lim }\limits_{x \to 0} \frac{{\tan x}}{x} = {\lim }\limits_{x \to 0} \frac{{{{\tan }^{ - 1}}x}}{x} = 1 {\lim }\limits_{x \to 0} \frac{{{{\sin }^{ - 1}}x}}{x} = 1

The above concept is used to solve limit problems involving trigonometric functions. We also use substitution, factorization, rationalization and other algebraic methods to evaluate these types of problems.

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Example:-

Example 4: Evaluate.

Ans:-       Because sec x = 1/cos x, you find that

 

 

L’ Hospital Rule:- L’Hôpital’s rule or L’Hospital’s rule (French: [lopital] was first introduced by  French mathematician Guillaume de l’Hôpital in his 1696 treatise, this is supposed to be the first book on Differential calculus. But the rule was originally discovered in 1694 by  Johann Bernoulli who was a Swiss Mathematician.

If f(x) and g(x) be two functions in such a way that

(i)   {\lim }\limits_{x \to a} f(x) = {\lim }\limits_{x \to a} f(x) = 0 or   {\lim }\limits_{x \to a} f(x) = {\lim }\limits_{x \to a} f(x) = \infty

 

(ii) both are continuous at x=a

 

(iii) both are differentiable at x=a

 

(iv) f'(x) and g'(x) are both continuous at x=a then

 

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

If the result is still in 0/0 or in the \frac{\infty }{\infty }form we can again differentiate and write like this

 

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

You can download this PDF to get Practice Questions. You can put your doubts in the comment section for a quick reply

 Mains_Limit_Paper-1.pdf

 Limits……Exercises.pdf

 Worksheets on Limits .pdf

 Worksheets on Limits.pdf

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