If limx-e2x-ex-x 1x-ea- then the value of a is

L=e^lim_x→∞ln(e^2x+e^x+x )x Using L Hospital's rule ⇒ e^lim_x→∞2e^2x+e^x+1e^2x+e^x+x=e^a Now, dividing by e^2x and since we know x→∞⇒ e^ 2x→ 0 So, we have nb ...

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Standard XII

Mathematics

Right Hand Derivative

If limx→∞e2x+...

Question

If $lim x \to \infty {(e^{2 x} + e^{x} + x)}^{\frac{1}{x}} = e^{a}$ , then the value of $a$ is

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Solution

$L = e^{lim x \to \infty \frac{ln (e^{2 x} + e^{x} + x)}{x}}$
Using L-Hospital's rule
$\Rightarrow e^{lim x \to \infty \frac{2 e^{2 x} + e^{x} + 1}{e^{2 x} + e^{x} + x}} = e^{a}$
Now, dividing by $e^{2 x}$ and since we know $x \to \infty \Rightarrow e^{- 2 x} \to 0$
So, we have $e^{a} = e^{⎛ ⎝ \frac{2 + 0 + 0}{1 + 0 + 0} ⎞ ⎠}$
$∴ e^{2} = e^{a} \Rightarrow a = 2$

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Right Hand Derivative

Standard XII Mathematics

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If limx→∞(e2x+ex+x)1x=ea, then the value of a is

L=elimx→∞ln(e2x+ex+x)x Using L-Hospital's rule ⇒elimx→∞2e2x+ex+1e2x+ex+x=ea Now, dividing by e2x and since we know x→∞⇒e−2x→0 So, we have ea=e⎛⎝2+0+01+0+0⎞⎠ ∴e2=ea ⇒a=2

If $lim x \to \infty {(e^{2 x} + e^{x} + x)}^{\frac{1}{x}} = e^{a}$ , then the value of $a$ is