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Limits At Infinity

Learn limits involving trigonometric functions, practice limits at positive & negative infinity and evaluation of more limits at infinity.

Limits at infinity

Limits at  positive infinity$$(+\infty)$$

• Let $$f$$be a function defined in some interval $$(a,\infty)$$ then

$$\lim\limits_{x\rightarrow\infty}\;f(x)=L$$ means that $$f(x)$$ can be made arbitrarily close to L by taking  $$x$$ sufficiently large.

• We say $$f(x)\rightarrow L$$ as $$x\rightarrow \infty$$.
• $$\infty$$ is not a number.  For the function whose graph is shown   $$\lim\limits_{x\rightarrow\infty}=f(x)=5$$.

Limits at minus infinity $$(-\infty)$$

• Let $$f$$be a function defined in some interval $$(-\infty,a)$$ then

$$\lim\limits_{x\rightarrow-\infty}\;f(x)=L$$ means that $$f(x)$$ can be made arbitrarily close to L by taking  $$x$$ sufficiently large negative.

• We say $$f(x)\rightarrow L$$ as $$x\rightarrow -\infty$$.

For the function $$f$$ whose graph is shown, find the value of $$\lim\limits_{x\rightarrow \infty}\;f(x)$$ .

A 2

B 5

C –2

D 1

×

$$\lim\limits_{x\rightarrow \infty}\;f(x)=$$ value to which $$f(x)$$ approaches as  $$x$$ takes very large positive values.

Graph of $$f$$approaches the limit $$y=2$$  as  $$x$$ take very large positive values.

$$\therefore \lim\limits_{x\rightarrow\infty}f(x)=2$$

For the function $$f$$ whose graph is shown, find the value of $$\lim\limits_{x\rightarrow \infty}\;f(x)$$ . A

2

.

B

5

C

–2

D

1

Option A is Correct

Theorem

• If $$r>0$$ is a rational number then $$\lim\limits_{x\to\infty} \dfrac{1}{x^r}=0$$.
• $$\lim\limits_{x\to-\infty} \dfrac{1}{x^r}=0$$ (if $$x^r$$ is defined)

Evaluate $$\lim\limits_{x\to\infty} \left(\dfrac{1}{x^3}\right)$$

A $$0$$

B $$-1$$

C $$\infty$$

D $$2$$

×

$$\lim\limits_{x\to\infty} \dfrac{1}{x^r}=0$$ for all  $$r>0$$

$$\therefore\lim\limits_{x\to\infty} \dfrac{1}{x^3}=0 \,\,\,\,(r=3)$$

Evaluate $$\lim\limits_{x\to\infty} \left(\dfrac{1}{x^3}\right)$$

A

$$0$$

.

B

$$-1$$

C

$$\infty$$

D

$$2$$

Option A is Correct

Evaluation of Limits at Infinity

• If we want to evaluate any limit of the form $$\lim \limits _{x\to\infty} \dfrac{f(x)}{g(x)}$$ where $$f(x),g(x)$$ are function containing powers of $$x$$ (not necessarily integer) then we look  for the highest power of $$x$$ .
1. If highest power of $$x$$ is present in $$f(x)$$ but not in $$g(x)$$ we say $$\lim \limits _{x\to\infty} \dfrac{f(x)}{g(x)} =\infty$$ (divide numerator and denominator  by that power of $$x$$)
2. If highest power of $$x$$ is present  in $$g(x)$$ and  not in $$f(x)$$ we say  $$\lim \limits _{x\to\infty} \dfrac{f(x)}{g(x)} =0$$  (divide numerator and denominator  by that power of $$x$$)
3. If highest  power  of $$x$$ is there in both $$f(x)$$ and  $$g(x)$$ then divide by that power of $$x$$ in numerator and denominator .

Evaluate  $$\lim\limits_{x\to\infty} \left(\dfrac{x^{5/2} +5x^2-3x+7}{x^3+8x+3}\right)$$

A $$\dfrac{1}{2}$$

B $$0$$

C $$\dfrac{5}{2}$$

D $$3$$

×

The highest power of $$x$$ in the denominator is $$x^3$$ , divide by $$x^3$$ in numerator and denominator.

$$= \lim \limits _{x\to\infty} \left(\dfrac{\left(\dfrac{x^{5/2}+5x^2-3x+7}{x^3}\right)}{\left(\dfrac{x^{3}+8x+3}{x^3}\right)}\right)$$

$$=\lim \limits _{x\to\infty} \left(\dfrac{{\dfrac{1}{x^{1/2}}+\dfrac{5}{x}-\dfrac{3}{x^2}+\dfrac{7}{x^3}}}{1+\dfrac{8}{x^2}+\dfrac{7}{x^3}}\right)$$

$$= \dfrac{\left(\lim \limits _{x\to\infty} \,\dfrac{1}{x^{1/_2}}\right) + \left(\lim \limits _{x\to\infty} \,\dfrac{5}{x}\right)- \left(\lim \limits _{x\to\infty} \,\dfrac{3}{x^{2}}\right)+ \left(\lim \limits _{x\to\infty} \,\dfrac{7}{x^{3}}\right)}{ \left(\lim \limits _{x\to\infty} \,1\right) +\left(\lim \limits _{x\to\infty} \,\dfrac{8}{x^{2}}\right)+ \left(\lim \limits _{x\to\infty} \,\dfrac{7}{x^{3}}\right)}$$

$$= \dfrac{0+0-0+0}{1+0+0} = \dfrac{0}{1} = 0$$

Evaluate  $$\lim\limits_{x\to\infty} \left(\dfrac{x^{5/2} +5x^2-3x+7}{x^3+8x+3}\right)$$

A

$$\dfrac{1}{2}$$

.

B

$$0$$

C

$$\dfrac{5}{2}$$

D

$$3$$

Option B is Correct

Limits of the Form infinity minus infinity$$(\infty-\infty)$$

• When limits of the form $$(\infty-\infty)$$ is given we first use algebra to create  $$\dfrac{\infty}{\infty}$$ form and that proceed  by dividing  an appropriate power of $$x$$ .
• Appropriate power mean the highest power  of $$x$$ that is occurring in numerator and denominator combined .

Evaluate   $$\lim\limits_{x\to\infty}\,\left(\sqrt{x^2+x+7}-\sqrt{x^2+2\,x+3}\right)$$

A $$\dfrac{4}{7}$$

B $$\dfrac{-1}{2}$$

C $$\dfrac{1}{2}$$

D $$\dfrac{5}{3}$$

×

$$\lim\limits_{x\to\infty}\,\left(\sqrt{x^2+x+7}-\sqrt{x^2+2\,x+3}\right) = \infty -\infty$$  form (note that both term are very large when $$x$$ is large)

Multiply the numerator and denominator by conjugate radical.

$$\lim\limits_{x\to\infty}\,\left(\sqrt{x^2+x+7}-\sqrt{x^2+2\,x+3}\right) \left(\sqrt{x^2+x+7}+\sqrt{x^2+2\,x+3}\right)$$

$$\left(\sqrt{x^2+x+7}+\sqrt{x^2+2\,x+3}\right) \left((u+v)(u-v)=u^2-v^2\right)$$

$$\Rightarrow\lim\limits_{x\to\infty}\dfrac{{x^2+x+7}-{x^2-2\,x-3}}{\sqrt{x^2+x+7}+\sqrt{x^2+2\,x+3}}$$

$$\Rightarrow\lim\limits_{x\to\infty}\,\dfrac{{(-x+4)}}{\sqrt{x^2+x+7}+\sqrt{x^2+2\,x+3}}$$

Divide numerator and denominator by $$x$$ :

$$\Rightarrow\lim\limits_{x\to\infty}\,\dfrac{{-1+\dfrac{4}{x}}}{\sqrt{1+\dfrac{1}{x}+\dfrac{7}{x^2}}+\sqrt{1+\dfrac{2}{x}+\dfrac{3}{x^2}}}$$

$$\Rightarrow\dfrac{-1+0}{\sqrt{1+0+0}+\sqrt{1+0+0}} = -\dfrac{1}{2}$$

Evaluate   $$\lim\limits_{x\to\infty}\,\left(\sqrt{x^2+x+7}-\sqrt{x^2+2\,x+3}\right)$$

A

$$\dfrac{4}{7}$$

.

B

$$\dfrac{-1}{2}$$

C

$$\dfrac{1}{2}$$

D

$$\dfrac{5}{3}$$

Option B is Correct

Limits at Positive Infinity

• To evaluate  $$\lim\limits_{x\to\infty} \dfrac{P(x)}{Q(x)}$$  where  $$P(x)$$ and $$Q(x)$$ are polynomials in $$x$$ , divide  the numerator and denominator  (i.e  $$P(x)$$ and $$Q(x)$$ by highest power of  $$x$$ that occur in $$Q(x)$$ ).
• Then use the property

$$\lim\limits_{x\to\infty} \dfrac{1}{x^r}=0$$      $$(r>0)$$ .

Evaluate     $$\lim\limits_{x\to\infty} \left(\dfrac{2x^2 +7x-6}{3x^2+8x+1}\right)$$

A $$\dfrac{3}{2}$$

B $$\dfrac{2}{3}$$

C $$\dfrac{7}{8}$$

D $$6$$

×

The highest power  in denominator $$Q(x)$$  is $$x^2$$ , divide by $$x^2$$ in Numerator and  Denominator.

$$=\lim\limits_{x\to\infty} \left[{\dfrac{\left(\dfrac{2x^2 +7x-6}{x^2}\right)}{\left(\dfrac{{3x^2+8x+1}}{x^2}\right)}}\right]$$

$$=\lim\limits_{x\to\infty} \left(\dfrac{2+\dfrac{7}{x}-\dfrac{6}{x^2}}{3+\dfrac{8}{x}+\dfrac{7}{x^2}}\right)$$

$$=\dfrac{\Bigg(\lim\limits_{x\to\infty}2\Bigg) +\Bigg(\lim\limits_{x\to\infty}\dfrac{7}{x}\Bigg) - \Bigg(\lim\limits_{x\to\infty}\dfrac{6}{x^2}\Bigg)}{\Bigg(\lim\limits_{x\to\infty}3\Bigg) +\Bigg(\lim\limits_{x\to\infty}\dfrac{8}{x}\Bigg) + \Bigg(\lim\limits_{x\to\infty}\dfrac{1}{x^2}\Bigg)}$$

$$= \dfrac{2+0-0}{3+0+0} = \dfrac{2}{3}$$

Evaluate     $$\lim\limits_{x\to\infty} \left(\dfrac{2x^2 +7x-6}{3x^2+8x+1}\right)$$

A

$$\dfrac{3}{2}$$

.

B

$$\dfrac{2}{3}$$

C

$$\dfrac{7}{8}$$

D

$$6$$

Option B is Correct

Limits at Negative Infinity

• To evaluate  $$\lim\limits_{x\to-\infty} \dfrac{P(x)}{Q(x)}$$  where  $$P(x)$$ and $$Q(x)$$ are polynomials in $$x$$ , divide  the numerator and denominator  (i.e  $$P(x)$$ and $$Q(x)$$ by highest power of  $$x$$ that occur in $$Q(x)$$ ).
• Then ,use the property $$\lim\limits_{x\to-\infty} \dfrac{1}{x^r}=0$$      $$(r>0)$$ .

Evaluate     $$\lim\limits_{x\to-\infty} \left(\dfrac{2x^2 +7x-6}{3x^2+8x+1}\right)$$ .

A $$\dfrac{3}{2}$$

B $$\dfrac{2}{3}$$

C $$\dfrac{7}{8}$$

D $$6$$

×

The highest power  in denominator $$Q(x)$$  is $$x^2$$ , divide by $$x^2$$ in Numerator and Denominator.

$$=\lim\limits_{x\to-\infty} \left[{\dfrac{\left(\dfrac{2x^2 +7x-6}{x^2}\right)}{\left(\dfrac{{3x^2+8x+1}}{x^2}\right)}}\right]$$

$$=\lim\limits_{x\to-\infty} \left(\dfrac{2+\dfrac{7}{x}-\dfrac{6}{x^2}}{3+\dfrac{8}{x}+\dfrac{1}{x^2}}\right)$$

$$=\dfrac{\Bigg(\lim\limits_{x\to-\infty}2\Bigg) +\Bigg(\lim\limits_{x\to-\infty}\dfrac{7}{x}\Bigg) - \Bigg(\lim\limits_{x\to-\infty}\dfrac{6}{x^2}\Bigg)}{\Bigg(\lim\limits_{x\to-\infty}3\Bigg) +\Bigg(\lim\limits_{x\to-\infty}\dfrac{8}{x}\Bigg) + \Bigg(\lim\limits_{x\to-\infty}\dfrac{1}{x^2}\Bigg)}$$

$$= \dfrac{2+0-0}{3+0+0} = \dfrac{2}{3}$$

Evaluate     $$\lim\limits_{x\to-\infty} \left(\dfrac{2x^2 +7x-6}{3x^2+8x+1}\right)$$ .

A

$$\dfrac{3}{2}$$

.

B

$$\dfrac{2}{3}$$

C

$$\dfrac{7}{8}$$

D

$$6$$

Option B is Correct

Infinite Limits At Infinity

• $$\lim\limits_{x\to\infty} \,f(x) = \infty$$ means that values of $$f(x)$$ becomes large positive  when  $$x$$ takes large positive values .
• $$\lim\limits_{x\to-\infty} \,f(x) = \infty$$ mean that values of $$f(x)$$ becomes large positive when  $$x$$ takes large negative values .
• $$\lim\limits_{x\to\infty} \,f(x) =- \infty$$ mean that values of $$f(x)$$ becomes large negative when  $$x$$ takes large positive values .

• $$\lim\limits_{x\to-\infty} \,f(x) =- \infty$$  mean that values of $$f(x)$$ becomes large negative when  $$x$$ takes large negative values .

Find   $$\lim\limits_{x\to\infty} \,(4x^3-x^5)$$ .

A $$\infty$$

B $$-\infty$$

C $$0$$

D $$\dfrac{1}{2}$$

×

$$\lim\limits_{x\to\infty} \,(4x^3-x^5) = \lim\limits_{x\to\infty} \,\overbrace {x^3}^{large +} (4-\overbrace{x^2}^{large - })$$

$$=\infty × -\infty = -\infty$$

Find   $$\lim\limits_{x\to\infty} \,(4x^3-x^5)$$ .

A

$$\infty$$

.

B

$$-\infty$$

C

$$0$$

D

$$\dfrac{1}{2}$$

Option B is Correct

Limits of Inverse Trigonometric Function

• Most of the limits which involve function having inverse trigonometric function  are evaluated just by the knowledge of the values of these function for various values of $$x$$
•  e.g. $$\lim\limits_{x\to0^+} (tan^{-1}(\ell n \,x)) = tan^{-1} (-\infty) = \dfrac{-\pi}{2}$$ because we know that  $$\ell n \,x \to- \infty$$ as $$x\to 0^+$$ .

Evaluate $$\lim\limits_{x\to\infty} sin^{-1} \left(\dfrac{2x^2 +5}{4x^2 +7}\right)$$

A $$\dfrac{\pi}{3}$$

B $$\dfrac{\pi}{6}$$

C $$\dfrac{\pi}{2}$$

D $$\dfrac{-\pi}{4}$$

×

$$\lim\limits_{x\to \infty} \dfrac{P(x)}{Q(x)} = \dfrac{{\text{Coefficient of highest power of x in P(x)}}}{{\text{Coefficient of highest power of x in Q(x)}}}$$

If  $${\text{degree }}P(x) = {\text{degree }}Q(x)$$

In this case $$\lim\limits_{x\to\infty} sin^{-1} \left(\dfrac{2x^2 +4}{4x^2 +7}\right)$$

$$\Rightarrow \lim\limits_{x\to\infty} sin^{-1} \left(\dfrac{2 +\dfrac{4}{x^2}}{4 +\dfrac{7}{x^2}}\right) = sin ^{-1} \dfrac{2}{4} = sin ^{-1} \dfrac{1}{2}$$

$$\Rightarrow\dfrac{\pi}{6}$$

Evaluate $$\lim\limits_{x\to\infty} sin^{-1} \left(\dfrac{2x^2 +5}{4x^2 +7}\right)$$

A

$$\dfrac{\pi}{3}$$

.

B

$$\dfrac{\pi}{6}$$

C

$$\dfrac{\pi}{2}$$

D

$$\dfrac{-\pi}{4}$$

Option B is Correct