##### Worked Examples – Limits

The indeterminate forms of limits

$\displaystyle \infty \cdot \infty ,\infty \cdot 0,\frac{\infty }{\infty },\frac{0}{0},\frac{\infty }{0},\infty +\infty ,\infty -\infty {{,1}^{\infty }}{{,0}^{\infty }},{{\infty }^{0}}$

The indeterminate forms of limits

$\displaystyle \infty \cdot \infty ,\infty \cdot 0,\frac{\infty }{\infty },\frac{0}{0},\frac{\infty }{0},$

$\displaystyle \infty +\infty ,\infty -\infty ,{{1}^{\infty }},{{0}^{\infty }},{{\infty }^{0}}$

### Important Limits

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{\sin x}}{x}=1$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{tgx}}{x}=1$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{{{e}^{x}}-1}}{x}=1$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{{{a}^{x}}-1}}{x}=\ln a$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{\ln (1+x)}}{x}=1$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{{{{\log }}_{a}}(1+x)}}{x}=\frac{1}{{\ln a}}$

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,{{(1+\frac{1}{x})}^{x}}=e$

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{{n-1}}}+…..{{a}_{{n-1}}}x+{{a}_{n}}}}{{{{b}_{0}}{{x}^{m}}+{{b}_{1}}{{x}^{{m-1}}}+…..{{b}_{{m-1}}}x+{{b}_{m}}}}=\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{a}_{0}}{{x}^{n}}}}{{{{b}_{0}}{{x}^{m}}}}~~(m,n\in N)$

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{a}_{0}}{{x}^{n}}+…..{{a}_{{n-1}}}x+{{a}_{n}}}}{{{{b}_{0}}{{x}^{m}}+…..{{b}_{{m-1}}}x+{{b}_{m}}}}$

$\displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{a}_{0}}{{x}^{n}}}}{{{{b}_{0}}{{x}^{m}}}}~~(m,n\in N)$

Example 1: Find the limits by Direct Substitution.

We simply substitute the value of x and evaluate.

a) $\displaystyle \underset{{x\to 1}}{\mathop{{\lim (}}}\,3{{x}^{2}}+2)$

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim (}}}\,3{{x}^{2}}+2)=\left[ {3{{{(1)}}^{2}}+2} \right]=$

$\displaystyle (3+2)=5$

b) $\displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,(-x-\sqrt{x})$

$\displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,(-x-\sqrt{x})=(-4-\sqrt{4})$

$\displaystyle=(-4-2)=-6$

c) $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,(3+\frac{2}{{x+4}})$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,(3+\frac{2}{{x+4}})=(3+\frac{2}{{0+4}})$

$\displaystyle=(3+\frac{2}{4})=(3+\frac{1}{2})$

$\displaystyle =\left( {\frac{{6+1}}{2}} \right)=\frac{7}{2}$

Example 2: Find the limits by Factoring.

If we just substitute the value as above we get the undefined form $\displaystyle \frac{0}{0}$

Wherever you see a quotient of polynomials you can try factoring if possible.

a) $\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}-4}}{{x-2}}=\underset{{x\to 2}}{\mathop{{\lim }}}\,\frac{{{{{(2)}}^{2}}-4}}{{2-2}}=\frac{0}{0}$

$\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}-4}}{{x-2}}$

We know that $\displaystyle {{a}^{2}}-{{b}^{2}}=(a-b)(a+b)$

So we can write $\displaystyle {{x}^{2}}-4=(x-2)(x+2)$

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

$\displaystyle =\underset{{x\to 2}}{\mathop{{\lim }}}\,(x+2)=(2+2)=4$

b) $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}-1}}{{x-1}}=\frac{{{{1}^{3}}-1}}{{1-1}}=\frac{0}{0}$

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}-1}}{{x-1}}$

We know that $\displaystyle {{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})$

So we can write $\displaystyle {{x}^{3}}-{{1}^{3}}=(x-1)({{x}^{2}}+x+1)$

$\displaystyle =\underset{{x\to 1}}{\mathop{{\lim }}}\,({{x}^{2}}+x+1)$

$\displaystyle =({{1}^{2}}+1+1)=3$

c) $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{1-x}}{{2x-2}}$

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{1-x}}{{2x-2}}=\frac{{1-1}}{{2-2}}=\frac{0}{0}$

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{1-x}}{{2x-2}}=\underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{-(x-1)}}{{2(x-1)}}$

$\displaystyle =\underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{-1}}{2}=-\frac{1}{2}$

Example 3: Find the limits by applying Rationalization.

If we just substitute the value as above we get an undefined form at all of them.
If we have a square root we multiply by the conjugate.

a) $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{1-\sqrt{x}}}{{1-x}}$

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{1-\sqrt{x}}}{{1-x}}=\frac{{1-\sqrt{1}}}{{1-1}}=\frac{0}{0}$

We know that$\displaystyle (\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=a-b$

So $\displaystyle (1-\sqrt{x})(1+\sqrt{x})=1-x$

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{1-\sqrt{x}}}{{1-x}}=\underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{(1-\sqrt{x})(1+\sqrt{x})}}{{(1-x)(1+\sqrt{x})}}$

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{(1-x)}}{{(1-x)(1+\sqrt{x})}}=\underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{1}{{(1+\sqrt{x})}}$

$\displaystyle =\underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{1}{{(1+\sqrt{1})}}=\frac{1}{2}$

a) $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{1-\sqrt{x}}}{{1-x}}$

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{1-\sqrt{x}}}{{1-x}}=\frac{{1-\sqrt{1}}}{{1-1}}=\frac{0}{0}$

We know that$\displaystyle (\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=a-b$

So $\displaystyle (1-\sqrt{x})(1+\sqrt{x})=1-x$

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{1-\sqrt{x}}}{{1-x}}$

$\displaystyle =\underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{(1-\sqrt{x})(1+\sqrt{x})}}{{(1-x)(1+\sqrt{x})}}$

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{(1-x)}}{{(1-x)(1+\sqrt{x})}}$

$\displaystyle =\underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{1}{{(1+\sqrt{x})}}$

$\displaystyle =\underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{1}{{(1+\sqrt{1})}}=\frac{1}{2}$

b) $\displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{{\sqrt{{x+6}}-3}}{{x-3}}$

$\displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{{\sqrt{{x+6}}-3}}{{x-3}}=\frac{{\sqrt{{3+6}}-3}}{{3-3}}=\frac{0}{0}$

By multiplying with the conjugate we get$\displaystyle (\sqrt{{x+6}}-3)(\sqrt{{x+6}}+3)=x-3$

$\displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{{(\sqrt{{x+6}}-3)(\sqrt{{x+6}}+3)}}{{(x-3)(\sqrt{{x+6}}+3)}}=$

$\displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{{x+6-9}}{{(x-3)(\sqrt{{x+6}}+3)}}=$

$\displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{{(x-3)}}{{(x-3)(\sqrt{{x+6}}+3)}}=$

$\displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{1}{{(\sqrt{{x+6}}+3)}}=$

$\displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{1}{{(\sqrt{{3+6}}+3)}}=\frac{1}{6}$

c) $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{\sqrt{{{{x}^{2}}+100}}-10}}{{{{x}^{2}}}}$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{\sqrt{{{{x}^{2}}+100}}-10}}{{{{x}^{2}}}}=\frac{{\sqrt{{0+100}}-10}}{0}=\frac{0}{0}$

By multiplying with the conjugate we get$\displaystyle (\sqrt{{{{x}^{2}}+100}}-10)(\sqrt{{{{x}^{2}}+100}}+10)={{x}^{2}}+100-100={{x}^{2}}$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{(\sqrt{{{{x}^{2}}+100}}-10)(\sqrt{{{{x}^{2}}+100}}+10)}}{{{{x}^{2}}(\sqrt{{{{x}^{2}}+100}}+10)}}=$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}}}{{{{x}^{2}}(\sqrt{{{{x}^{2}}+100}}+10)}}=$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{1}{{(\sqrt{{{{x}^{2}}+100}}+10)}}=$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{1}{{(\sqrt{{{{0}^{2}}+100}}+10)}}=\frac{1}{{20}}$

c) $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{\sqrt{{{{x}^{2}}+100}}-10}}{{{{x}^{2}}}}$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{\sqrt{{{{x}^{2}}+100}}-10}}{{{{x}^{2}}}}$

$\displaystyle =\frac{{\sqrt{{0+100}}-10}}{0}=\frac{0}{0}$

By multiplying with the conjugate we get

$\displaystyle (\sqrt{{{{x}^{2}}+100}}-10)(\sqrt{{{{x}^{2}}+100}}+10)$

$\displaystyle ={{x}^{2}}+100-100={{x}^{2}}$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{(\sqrt{{{{x}^{2}}+100}}-10)(\sqrt{{{{x}^{2}}+100}}+10)}}{{{{x}^{2}}(\sqrt{{{{x}^{2}}+100}}+10)}}=$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}}}{{{{x}^{2}}(\sqrt{{{{x}^{2}}+100}}+10)}}=$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{1}{{(\sqrt{{{{x}^{2}}+100}}+10)}}=$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{1}{{(\sqrt{{{{0}^{2}}+100}}+10)}}=\frac{1}{{20}}$

Example 4: Find the limits that goes to infinity.

What we should keep in mind is that $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{1}{x}=0$

a) $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2{{x}^{3}}+3x-1}}{{{{x}^{3}}+2}}$

If we find the limit by substituting we see that we get the undefined form $\displaystyle \frac{\infty }{\infty }$

We can divide with the highest common factor or factorise the highest common factor.

Dividing with x3:

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\frac{{2{{x}^{3}}}}{{{{x}^{3}}}}+\frac{{3x}}{{{{x}^{3}}}}-\frac{1}{{{{x}^{3}}}}}}{{\frac{{{{x}^{3}}}}{{{{x}^{3}}}}+\frac{2}{{{{x}^{3}}}}}}=$

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2+\frac{3}{{{{x}^{2}}}}-\frac{1}{{{{x}^{3}}}}}}{{1+\frac{2}{{{{x}^{3}}}}}}=$

We know that $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{3}{{{{x}^{2}}}}=0$

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{1}{{{{x}^{3}}}}=0$

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{2}{{{{x}^{3}}}}=0$

So $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2+0-0}}{{1+0}}=2$

b) $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{3{{x}^{5}}+2{{x}^{3}}+1}}{{9{{x}^{5}}+12x+3}}$

This is the undefined form $\displaystyle \frac{\infty }{\infty }$

Firstly we will see which term is dominating on the denominator and numerator

The terms are $\displaystyle {3{{x}^{5}}}$ and $\displaystyle {9{{x}^{5}}}$

As we approach infinity we get: $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{3{{x}^{5}}+2{{x}^{3}}+1}}{{9{{x}^{5}}+12x+3}}=$

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{3{{x}^{5}}}}{{9{{x}^{5}}}}=\frac{3}{9}=\frac{1}{3}$

Example 5: Find the limits.

a) $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{10{{x}^{{10}}}+{{{10}}^{{10}}}}}{{{{{(10{{x}^{2}}+5)}}^{5}}}}$

We have the undefined form $\displaystyle \frac{\infty }{\infty }$

We apply the ratio of the highest terms up and down the line.

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{10{{x}^{{10}}}+{{{10}}^{{10}}}}}{{{{{(10{{x}^{2}}+5)}}^{5}}}}=$

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{10{{x}^{{10}}}}}{{{{{10}}^{5}}{{x}^{{10}}}}}=\frac{1}{{10}}$

b) $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{\sqrt{{x+3}}-2}}{{{{x}^{2}}-1}}$

We have the undefined form $\displaystyle \frac{0}{0}$

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{\sqrt{{x+3}}-2}}{{{{x}^{2}}-1}}=$

We multiply with the conjugate $\displaystyle {(\sqrt{{x+3}}+2)}$

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{(\sqrt{{x+3}}-2)(\sqrt{{x+3}}+2)}}{{({{x}^{2}}-1)(\sqrt{{x+3}}+2)}}=$

We know that $\displaystyle (a-b)(a+b)={{a}^{2}}-{{b}^{2}}$

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{x+3-4}}{{({{x}^{2}}-1)(\sqrt{{x+3}}+2)}}=$

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{x-1}}{{(x-1)(x+1)(\sqrt{{x+3}}+2)}}=$

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{1}{{(x+1)(\sqrt{{x+3}}+2)}}=$

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{1}{{(1+1)(\sqrt{{1+3}}+2)}}=\frac{1}{8}$

c) $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{x+\sin 3x}}{{x-\sin 2x}}$

We have the undefined form $\displaystyle \frac{0}{0}$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{x+\sin 3x}}{{x-\sin 2x}}=$

Factorise one x.

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{x(1+\frac{{\sin 3x}}{x})}}{{x(1-\frac{{\sin 2x}}{x})}}=$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{(1+\frac{{\sin 3x}}{x})}}{{(1-\frac{{\sin 2x}}{x})}}=$

We multiply and divide each of sinuses with 3 and 2 to form the limit of an important limit.

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{(1+3\frac{{\sin 3x}}{{3x}})}}{{(1-2\frac{{\sin 2x}}{{2x}})}}=$

We substitute 3x=t and 2x=u

$\displaystyle \frac{{\underset{{t\to 0}}{\mathop{{\lim }}}\,(1+3\frac{{\sin t}}{t})}}{{\underset{{u\to 0}}{\mathop{{\lim }}}\,(1-2\frac{{\sin u}}{u})}}=$

$\displaystyle \frac{{\underset{{t\to 0}}{\mathop{{\lim }}}\,(1+3\frac{{\sin t}}{t})}}{{\underset{{u\to 0}}{\mathop{{\lim }}}\,(1-2\frac{{\sin u}}{u})}}=\frac{{1+3}}{{1-2}}=-4$

d) $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{{{3}^{{2x+3}}}-27}}{{{{3}^{{x+1}}}-3}}$

We have the undefined form $\displaystyle \frac{0}{0}$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{{{3}^{{2x+3}}}-27}}{{{{3}^{{x+1}}}-3}}=$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{{{3}^{3}}({{3}^{{2x}}}-1)}}{{3({{3}^{x}}-1)}}=$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{{{3}^{2}}({{3}^{x}}-1)({{3}^{x}}+1)}}{{({{3}^{x}}-1)}}=$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,{{3}^{2}}({{3}^{x}}+1)=18$

e) $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,{{(1+\frac{5}{x})}^{x}}$)

We have the undefined form $\displaystyle {{1}^{\infty }}$

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,{{(1+\frac{5}{x})}^{x}}=$

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,{{\left[ {{{{(1+\frac{5}{x})}}^{{\frac{x}{5}}}}} \right]}^{5}}=$

We substitute t=x/5 and from a known limit we get:

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,{{\left[ {{{{(1+\frac{1}{t})}}^{t}}} \right]}^{5}}={{e}^{5}}$