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We use the definition of e^x to solve this problem.

e^x =  \lim_{n \to \infty} (1+\frac{x}{n})^n\\\\= 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!}+....\infty\\\\e^x=1 + x(1+\frac{x}{2!}+\frac{x^2}{3!}+....+\frac{x^{n-1}}{n!}+....\infty)\\\\1)\\ \lim_{h \to 0}\ \frac{e^{tan\ h}-1}{h} = \lim_{h \to 0}\ \frac{1}{h}[ 1+tan\ h(1+\frac{tan\ h}{2!}+\frac{tan^2\ h}{3!}+....\infty) - 1]\\\\=\lim_{h \to 0}\ \frac{tan\ h}{h}[ 1+\frac{tan\ h}{2!}+\frac{tan^2\ h}{3!}+....\infty]\\\\=1*\lim_{h \to 0}\ 1+\frac{tan\ h}{2!}+\frac{tan^2\ h}{3!}+....\infty\\\\=1\\

we use the following limits.

\lim_{h \to 0},\ tan\ h \to 0\ \ and\ \ \frac{tan\ h}{h} \to\ 1.

\lim_{h \to 0} \frac{e^{(x+h)^2}-e^{x^2}}{h}\\\\= \lim_{h \to 0} \frac{1}{h}*[e^{x^2+2xh+h^2}-e^{x^2}]\\\\=e^{x^2}*\lim_{h \to 0} \frac{1}{h}*[e^{2xh+h^2}-1]\\\\=e^{x^2}*\lim_{h \to 0} \frac{1}{h}*[1+2xh+h^2+\frac{(2xh+h^2)^2}{2!}+\frac{(2xh+h^2)^3}{3!}+....\infty\ \ -1]\\\\=e^{x^2}*\lim_{h \to 0} \frac{1}{h}*[2xh+h^2+\frac{(2xh+h^2)^2}{2!}+\frac{(2xh+h^2)^3}{3!}+....\infty]\\\\=e^{x^2}*\lim_{h \to 0} [2x+h+\frac{h(2x+h)^2}{2!}+\frac{h^2(2x+h)^3}{3!}+....\infty]\\\\=e^{x^2}*2x\\

the expression on the RHS simplifies as all the terms except tend to zero, as they contain h.

1 5 1
sir is there any other easier meathod its a 1 mark question.
take limit of h ->0 for derivative of numerator divided by derivative of denominator.
1) lim h->0 e^tan h * sec^2 h / 1 => e^0 * sec^2 0 = 1
Another way: given limit is the definition of e^tanh at h = 0. So evaluate its derivative and substitute h = 0.
2) same as above two methods.. given problem is same as the definition of the derivative of e^{x^2}. wrt x. so find it directly.
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