Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 110988 by Rio Michael last updated on 01/Sep/20

 Evaluate without using L′hopital′s rule    lim_(x→4)  (((√x)−2)/(x−4))

$$\:\mathrm{Evaluate}\:\mathrm{without}\:\mathrm{using}\:\mathrm{L}'\mathrm{hopital}'\mathrm{s}\:\mathrm{rule} \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{\sqrt{{x}}−\mathrm{2}}{{x}−\mathrm{4}} \\ $$

Answered by Rasheed.Sindhi last updated on 01/Sep/20

  lim_(x→4)  (((√x)−2)/(x−4))     (((√x)−2)/( ((√x))^2 −(2)^2 ))=(((√x)−2)/( ((√x)−2)((√x)+2)))  =(1/( ((√x)+2)))    lim_(x→4)  (((√x)−2)/(x−4))=lim_(x→4) (1/( ((√x)+2)))=(1/( (√4)+2))  =(1/4)

$$\:\:\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{\sqrt{{x}}−\mathrm{2}}{{x}−\mathrm{4}} \\ $$$$\:\:\:\frac{\sqrt{{x}}−\mathrm{2}}{\:\left(\sqrt{{x}}\right)^{\mathrm{2}} −\left(\mathrm{2}\right)^{\mathrm{2}} }=\frac{\sqrt{{x}}−\mathrm{2}}{\:\left(\sqrt{{x}}−\mathrm{2}\right)\left(\sqrt{{x}}+\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\:\left(\sqrt{{x}}+\mathrm{2}\right)} \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{\sqrt{{x}}−\mathrm{2}}{{x}−\mathrm{4}}=\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\frac{\mathrm{1}}{\:\left(\sqrt{{x}}+\mathrm{2}\right)}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}}+\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$

Commented by Rio Michael last updated on 01/Sep/20

correct sir.

$$\mathrm{correct}\:\mathrm{sir}. \\ $$

Answered by malwan last updated on 02/Sep/20

lim_(x→4) (((√x)−2)/(x−4))×(((√x)+2)/( (√x)+2))  =lim_(x→4) ((x−4)/((x−4)((√x)+2)))  =lim_(x→4) (1/( (√x)+2)) = (1/( (√4)+2)) = (1/4)

$$\underset{{x}\rightarrow\mathrm{4}} {{lim}}\frac{\sqrt{{x}}−\mathrm{2}}{{x}−\mathrm{4}}×\frac{\sqrt{{x}}+\mathrm{2}}{\:\sqrt{{x}}+\mathrm{2}} \\ $$$$=\underset{{x}\rightarrow\mathrm{4}} {{lim}}\frac{{x}−\mathrm{4}}{\left({x}−\mathrm{4}\right)\left(\sqrt{{x}}+\mathrm{2}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{4}} {{lim}}\frac{\mathrm{1}}{\:\sqrt{{x}}+\mathrm{2}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}}+\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Answered by bemath last updated on 01/Sep/20

let (√x) = b ⇒x = b^2   lim_(b→2)  ((b−2)/(b^2 −4)) = lim_(b→2)  (1/(b+2)) = (1/4)

$$\mathrm{let}\:\sqrt{\mathrm{x}}\:=\:\mathrm{b}\:\Rightarrow\mathrm{x}\:=\:\mathrm{b}^{\mathrm{2}} \\ $$$$\underset{\mathrm{b}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{b}−\mathrm{2}}{\mathrm{b}^{\mathrm{2}} −\mathrm{4}}\:=\:\underset{\mathrm{b}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{b}+\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com