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}$
TIP! See also Examples with Derivatives
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}$
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}}$
