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Chapter 1: Problem 5

$\lim _{x \rightarrow \infty}\left(1+a^{2}\right)^{x} \cdot \frac{b}{\left(1+a^{2}\right)^{x}}\( is \)(a, b \in R)$ (A) \(\sqrt{b}\) (B) b (C) \(\mathrm{b}^{2}\) (D) none of these

Short Answer

Expert verified
Question: Find the limit of the given function as \(x \rightarrow \infty\): \(\lim _{x \rightarrow \infty}\left((1+a^{2})^{x} \cdot \frac{b}{(1+a^{2})^{x}}\right)\) Options: A) \(\sqrt{b}\) B) b C) \({b}^{2}\) D) none of these Answer: (B) b

Step by step solution

01

Write down the given function

The given function is: \(\lim _{x \rightarrow \infty}\left((1+a^{2})^{x} \cdot \frac{b}{(1+a^{2})^{x}}\right)\)
02

Simplify the given function

We can simplify the given function by dividing the base of the power function in the numerator and denominator. \(\lim _{x \rightarrow \infty}\left(\frac{(1+a^{2})^{x} \cdot b}{(1+a^{2})^{x}}\right)\) By simplifying: \(\lim _{x \rightarrow \infty} \left(b\right)\)
03

Determine the limit

Since b is a constant, the limit is itself as x approaches infinity: \(\lim _{x \rightarrow \infty} \left(b\right) = b\)
04

Choose the correct option

By comparing the result to the given options, we see that our result matches option (B): Answer: (B) b

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Most popular questions from this chapter

$\lim _{x \rightarrow 0} \frac{6 x^{2}(\cot x)(\operatorname{cosec} 2 x)}{\sec \left(\cos x+\pi \tan \left(\frac{\pi}{4 \sec x}\right)-1\right)}$ has the value equal to (A) 6 (B) \(-6\) (C) 0 (D) \(-3\)

If \([.]\) represents greatest integer function, then which of the following, is identical to \(\mathrm{f}\) of \((\mathrm{x})\) for \(\mathrm{x} \in(\pi, 2 \pi)\) (A) \(\operatorname{sgn}(\mathrm{x})\) (B) \(\frac{x}{x}\) (C) \(\sin \pi[\mathrm{x}]\) (D) \(\frac{\sin x}{\sin x}\)

The true statement(s) is / are (A) If \(\mathrm{f}(\mathrm{x})<\mathrm{g}(\mathrm{x})\) for all $\mathrm{x} \neq \mathrm{a}\(, then \)\lim _{\mathrm{x} \rightarrow \mathrm{a}} \mathrm{f}(\mathrm{x})<\lim _{x \rightarrow a} \mathrm{~g}(\mathrm{x})$. (B) If \(\lim _{x \rightarrow c} \mathrm{f}(x)=0\) and \(|g(x)| \leq M\) for a fixed number \(M\) and all \(x \neq c\), then \(\lim f(x) \cdot g(x)=0\) (C) If \(\lim _{x \rightarrow c} \mathrm{f}(\mathrm{x})=\mathrm{L}\), then $\lim _{\mathrm{x} \rightarrow \mathrm{c}}|\mathrm{f}(\mathrm{x})|=|\mathrm{L}|$ and conversely if \(\lim |\mathrm{f}(\mathrm{x})|=|\mathrm{L}|\) then $\lim _{x \rightarrow \infty} \mathrm{f}(\mathrm{x})=\mathrm{L}$. (D) If \(\mathrm{f}(\mathrm{x})=\mathrm{g}(\mathrm{x})\) for all real number other then \(\mathrm{x}=0\) and \(\lim _{x \rightarrow 0} f(x)=L\), then $\lim _{x \rightarrow 0} g(x)=L$

Let $\mathrm{f}(\mathrm{x})=\lim _{\mathrm{n} \rightarrow \infty} \frac{2 \mathrm{x}^{2 \mathrm{n}} \sin \frac{1}{\mathrm{x}}+\mathrm{x}}{1+\mathrm{x}^{2 \mathrm{n}}}$ then which of the following alternative(s) is/are correct ? (A) \(\lim _{x \rightarrow \infty} x f(x)=2\) (B) \(\lim \mathrm{f}(\mathrm{x})\) does not exist (C) \(\lim _{x \rightarrow 0} f(x)\) does not exist (D) \(\lim _{x \rightarrow-\gamma} \mathrm{f}(\mathrm{x})\) is equal to zero.

Let \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are non zero constant number then $\lim _{\mathrm{r} \rightarrow \infty} \frac{\cos \frac{\mathrm{a}}{\mathrm{r}}-\cos \frac{\mathrm{b}}{\mathrm{r}} \cos \frac{\mathrm{c}}{\mathrm{r}}}{\sin \frac{\mathrm{b}}{\mathrm{r}} \sin \frac{\mathrm{c}}{\mathrm{r}}}$ equals (A) \(\frac{a^{2}+b^{2}-c^{2}}{2 b c}\) (B) \(\frac{\mathrm{c}^{2}+\mathrm{a}^{2}-\mathrm{b}^{2}}{2 \mathrm{bc}}\) (C) \(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\) (D) independent of \(a, b, c\)
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