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Evaluate given limit: \lim _{x \rightarrow \infty}\left(\sqrt{\left(25 x^2+13 x+2\right)}-\sqrt{\left(25 x^2+2\right)}\right)

Option: 1

1


Option: 2

1.5


Option: 3

1.3


Option: 4

2


Answers (1)

best_answer

For solving the limits of the type ∞ - ∞, we simply rationalize the given limit, that is multiplying and dividing the limit by its additive inverse.

Additive inverse of \lim _{x \rightarrow \infty}\left(\sqrt{\left(25 x^2+13 x+2\right)}-\sqrt{\left(25 x^2+2\right)}\right) is \lim _{x \rightarrow \infty}\left(\sqrt{\left(25 x^2+13 x+2\right)}+\sqrt{\left(25 x^2+2\right)}\right)

Hence,

\begin{aligned} & =\lim _{x \rightarrow \infty} \frac{\left(\sqrt{\left(25 x^2+13 x+2\right)}-\sqrt{\left(25 x^2+2\right)}\right)\left(\sqrt{\left(25 x^2+13 x+2\right)}+\sqrt{\left(25 x^2+2\right)}\right)}{\left(\sqrt{\left(25 x^2+13 x+2\right)}+\sqrt{\left(25 x^2+2\right)}\right)} \\ & =\lim _{x \rightarrow \infty} \frac{\left(\left(25 x^2+13 x+2\right)-\left(25 x^2+2\right)\right)}{\left(\sqrt{\left(25 x^2+13 x+2\right)}+\sqrt{\left(25 x^2+2\right)}\right)} \\ & =\lim _{x \rightarrow \infty} \frac{13 x}{\left(\sqrt{\left(25 x^2+13 x+2\right)}+\sqrt{\left(25 x^2+2\right)}\right)} \\ & \end{aligned}\text { Dividing the numerator and denominator by } \mathrm{x} \text { : }

\begin{aligned} & =\lim _{x \rightarrow \infty} \frac{13}{\left(\sqrt{\left(25+\frac{13}{x}+\frac{2}{x^2}\right)}+\sqrt{\left.\left(25+\frac{2}{x^2}\right)\right)}\right.} \\ & =\frac{13}{2 \times 5} \\ & =\frac{13}{10} \\ & =1.3 \end{aligned}

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HARSH KANKARIA

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