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Question Number 44441 by Tawa1 last updated on 29/Sep/18

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Sep/18

lim_(t→0 )  ((e^(−5t) −1)/(−5t))×−5  =1×−5=−5  lim_(t→0) (t^(11) /(t^(11) (t−1)))=lim_(t→0) (1/(t−1))=−1  lim_(t→0) (1/(11+t))=(1/(11))

$$\underset{{t}\rightarrow\mathrm{0}\:} {\mathrm{lim}}\:\frac{{e}^{−\mathrm{5}{t}} −\mathrm{1}}{−\mathrm{5}{t}}×−\mathrm{5} \\ $$$$=\mathrm{1}×−\mathrm{5}=−\mathrm{5} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{t}^{\mathrm{11}} }{{t}^{\mathrm{11}} \left({t}−\mathrm{1}\right)}=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{t}−\mathrm{1}}=−\mathrm{1} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{11}+{t}}=\frac{\mathrm{1}}{\mathrm{11}} \\ $$

Commented by Tawa1 last updated on 29/Sep/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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