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  • Evaluate the limit <img alt="\mathrm{\lim _{x \rightarrow \infty} x\left(16 x^4+x^2+1\right)^{1 / 4}-2 x^2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crig

Evaluate the limit \mathrm{\lim _{x \rightarrow \infty} x\left(16 x^4+x^2+1\right)^{1 / 4}-2 x^2}

 

Option: 1

\frac{1}{32}


Option: 2

0


Option: 3

1


Option: 4

\frac{3}{5}


Answers (1)

best_answer

\mathrm{ \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1} }           (1)

Putting \mathrm{ x^2=1 / t } we can see that \mathrm{ t \rightarrow 0^{+} }and the expression under limit is transformed into

\mathrm{ 2 \cdot \frac{\left(1+t / 16+t^2 / 16\right)^{1 / 4}-1}{t}=2 \cdot \frac{\left(1+t / 16+t^2 / 16\right)^{1 / 4}-1}{t / 16+t^2 / 16} \cdot\left(\frac{1}{16}+\frac{t}{16}\right) }
and therefore the limit is equal to the limit of

\mathrm{ \frac{1}{8} \cdot \frac{\left(1+t / 16+t^2 / 16\right)^{1 / 4}-1}{t / 16+t^2 / 16} }

\mathrm{ as \: t \rightarrow 0^{+}. Next \: we \: put \: u=1+t / 16+t^2 / 16 \: so \: that \: \: u \rightarrow 1^{+} }and the desired limit is equal to

\mathrm{ \frac{1}{8} \lim _{u \rightarrow 1^{+}} \frac{u^{1 / 4}-1^{1 / 4}}{u-1} }

which equals \mathrm{ (1 / 8)(1 / 4)=1 / 32 } via formula (1).

Hence option 1 is correct.

Posted by

vinayak

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