LIMIT

Question

$\lim_{X\rightarrow \infty } -(x+1)(e^{\frac{1}{x+1}}-1)$

Answer 1

it is a indeterminant form

 $\lim_{X\rightarrow \infty } -(x+1)(e^{\frac{1}{x+1}}-1)$

converting into $\frac{0}{0}$ indeterminant form
$\lim_{X\rightarrow \infty } -\frac{(e^{\frac{1}{x+1}}-1)}{\frac{1}{(x+1)}}$

now this is in indeterminant form $\frac{0}{0}$ form so Lhosptal rule

differentiating num and denominator

$\lim_{X\rightarrow \infty } -\frac{(e^{\frac{1}{x+1}}*-1*\frac{1}{(x+1)^2})}{-1*\frac{1}{(x+1)^2}}$

so it will be

-$\lim_{X\rightarrow \infty } e^{\frac{1}{x+1}}$ by substituting we will get ans as -1

Comments

-1 correct it..you forgot minus

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yeah -1 forgot the sign

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even we could substitute $\frac{1}{x+1} = m$ sothat it becomes

$\lim_{m\rightarrow 0} \frac{e^{m}-1}{m}$

 

This is straighforward.

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yes nice!